Answer:
Explanation:
Sample Response: If global warming continues, corals will continue to expel the algae from their cells to avoid poisonous buildup. This will cause corals to die. Without corals, the algae are not protected and cannot perform photosynthesis. This will cause the algae to die as well.
How many atoms are in each elemental sample?
16.8 g Sr
26.5 g Fe
8.94 g Bi
40.0 g P
Explanation:
The number of atoms in 1mol of every element can be represented by Avogadro's number, which is [tex]6.022*10^{23}[/tex].
Knowing this, now we can find the atoms in each of these molecules!
[tex]16.8gSr*\frac{1molSr}{87.62gSr} *\frac{6.022*10^{23}atomsSr}{1molSr} =[/tex]
1.15*10^23 atoms of Sr
[tex]26.5gFe*\frac{1molFe}{55.85gFe} *\frac{6.022*10^{23}atomsFe}{1molFe} =[/tex]
2.86*10^23 atoms of Fe
[tex]8.94gBi*\frac{1molBi}{208.98gBi} *\frac{6.022*10^{23}atomsBi}{1molBi} =[/tex]
2.58*10^22 atoms of Bi
[tex]40.0gP*\frac{1molP}{30.97gP} *\frac{6.022*10^{23}atomsP}{1molP}=[/tex]
7.78*10^23 atoms of P
Consider the synthesis of water as shown in Model 3. A container is filled with 10,0 g of H, and
5.0 g of Oz
Which reactant (hydrogen or oxygen) is the limiting reactant in this case?
Answer:
Oxygen, O₂ is the limiting reactant
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2H₂ + O₂ —> 2H₂O
Next, we shall determine the masses of H₂ and O₂ that reacted from the balanced equation. This can be obtained as follow:
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass of H₂ from the balanced equation = 2 × 2 = 4 g
Molar mass of O₂ = 16 × 2 = 32 g/mol
Mass of O₂O from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂.
Finally, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂.
Therefore, 10 g of H₂ will react with
= (10 × 32)/4 = 80 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 80 g) of O₂ than what was given (i.e 5 g) is required to react completely with 10 g of H₂. Therefore, O₂ is the limiting reactant.
Oxygen has been the limiting reactant in the reaction.
A limiting reactant can be defined as the reactant in the reaction in which the product concentration has been dependent.
The balanced equation for the formation of water has been:
[tex]\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O[/tex]
For the formation of reaction to form 2 moles of water, 2 moles of hydrogen reacts with 1 mole of oxygen.
The moles can be calculated as:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
The moles of Hydrogen in 10 g [tex]\rm H_2[/tex]:
Moles = [tex]\rm \dfrac{10}{2}[/tex]
Moles of hydrogen = 5 mol.
Moles of Oxygen in 5 grams Oxygen:
Moles = [tex]\rm \dfrac{5}{32}[/tex]
Moles of oxygen = 0.156 mol.
For the reaction with 2 moles of Hydrogen 1 mole of Oxygen has been required.
For reacting with 5 mol of Hydrogen, moles of oxygen required are:
Moles of oxygen = [tex]\rm \dfrac{1}{2}\;\times\;5[/tex]
Moles of oxygen required = 2.5 moles.
The available oxygen = 0.156 moles.
Since the moles of oxygen available is lesser than required, the formation of the product has been dependent on the concentration of the oxygen.
Thus, oxygen has been the limiting reactant in the reaction.
For more information about the limiting reactant, refer to the link:
https://brainly.com/question/14225536
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
element & mass %
phosphorus & 39.18%
sulfur & 60.82%
Write the molecular formula of X.
Answer:
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
element & mass %
phosphorus & 39.18%
sulfur & 60.82%
Write the molecular formula of X.
Explanation:
The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.
Empirical formula calculation:
element: phosphorus sulfur
co9mposition: 39.185% 60.82%
divide with
atomic mass: 39.185/31.0 g/mol 60.82/32.0g/mol
=1.26mol 1.90mol
smallest mole ratio: 1.26mol/1.26mol =1 1.90mol/1.26 mol =1.50
multiply with 2: 2 3
Hence, the empirical formula is:
P2S3.
Mass of empirical formula is:
158.0g/mol
Given, molecule has molar mass --- 316.25 g/mol
Hence, the ratio is:
316.25g/mol/158.0 =2
Hence, the molecular formula of the compound is :
2 x (P2S3)
=[tex]P_4S_6[/tex]
Indicate if the following are the correct ground state electron configurations
for the atom listed by choosing correct or incorrect from the drop down menu.
1. Cr: [Ar]4s03d6
2. Zr: [Kr]5s23f144d2
3. Fe: [Ar]4s23d6
4. Co3+: [Ar]4s03d6
5. Ti2+: [Ar]4s03d2
6. Cu+: [Ar]4s23d8
Answer:
1) incorrect
2) incorrect
3) correct
4) correct
5) correct
6) incorrect
Explanation:
The correct electronic configuration of chromium is; [Ar] 3d⁵ 4s¹
The correct electronic configuration for Zr is; [Kr] 4d² 5s²
The correct electronic configuration of Cu^+ is; [Ar] 3d¹⁰
The electronic configuration of an atom refers to the arrangement of electrons in the atoms of such element.
The appropriate number of electrons and its properly written electronic configuration is clearly shown in this answer.
Ammonium phosphate is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid with liquid ammonia. Calculate the moles of phosphoric acid needed to produce 1.80 mol of ammonium phosphate. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Answer:
Explanation:
The reaction is
H3Po4+3NH3\to→ (NH4)3PO4
Given,7.10g NH3=7.10g/molar mass of NH3
=7.10g/(17.031g/mol)
=0.416mol
From the reaction
3 mol ammonia reacted and produced 1 mole of ammoniam phosphate
So,One mole ammonia reacted and produced 1/ 3 mole ammonium phosphate.
And Also,0.416 mole ammonium reacted and produced (1/3)0.416=0.138 mole ammonium phosphate.
Hence 0.138mole=0.138mole*149.08 g/mole
=20.573gm ammonium phosphate produced.
Hence 20.573g of ammonium phosphate is produced by the reaction of 7.10 g of ammonia.
A molecule with the formula AX 4 uses ________ to form its bonds. sp2 hybrid orbitals sp3d hybrid orbitals sp3 hybrid orbitals sp3d2 hybrid orbitals sp hybrid orbitals
Answer:
sp3 hybrid orbitals
Explanation:
The formula of a molecule gives us an idea of its structure and the nature of hybrid orbitals that are involved in the formation of the molecule.
AX4 corresponds to tetrahedral geometry. If a molecule is in tetrahedral geometry, it is most likely sp3 hybridized as usual.
Hence, a molecule with the formula AX 4 uses sp3 hybrid orbitals to form its bonds
How many moles are in the number of molecules below? I only need to know the 5th question.
Answer:
11
1. 6.02×10 23
this is the answer Hope it helps you
one mole of a perfect gas at 300K as an initial pressure at 15 atm and is allowed to contract isothermally to a pressure of 1atm. calculate the entropy change from this contraction
Answer:
-46.67 J.
Explanation:
We are given;
Initial Pressure = 15atm = 15 × 10^(3) J
Final pressure = 1atm = 1 × 10^(3) J
Temperature = 300k
The pressures were converted to Joules.
Formula for the entropy change is;
∆S_system = ∆S_surrounding = -(dQ)/T
-(dQ)/T = (-(15 × 10^(3)) - (1 × 10^(3))/300)
= -46.67 J.
Methyl orange can change color by transitioning from one chromophore to another. When added to a clear solution and the solution turns red, it is determined to be a(n) __________ in its __________ stable form.
Answer:
acidic titration in its stable form
Explanation:
Methyl orange can change its color in titration solution. The yellow color is towards alkaline solution and red color is towards acidic solution. The Ph value of solution will change during this chemical process.
A sample of hydrogen nitrate or nitric acid, HNO 3 contains 18.8 x 1022 molecules.
How much mass of nitric acid are in the sample?
Answer:
19.7 g.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to realize this problem can be solved by using a molecules-moles-mass relationship, starting with the given molecules, using the Avogadro's number and the molar mass of nitric acid (63.01 g/mol):
[tex]18.8x10^{22}molec*\frac{1mol}{6.022x10^{23}molec}* \frac{63.01g}{1mol} \\\\=19.7g[/tex]
Regards!
1. A 225-L barrel of white wine has an initial free SO2 concentration of 22 ppm and a pH of 3.70. How much SO2 (in grams) should be added to the barrel to result in the required SO2 level
Answer:
The appropriate answer is "9.225 g".
Explanation:
Given:
Required level,
= 63 ppm
Initial concentration,
= 22 ppm
Now,
The amount of free SO₂ will be:
= [tex]Required \ level -Initial \ concentration[/tex]
= [tex]63-22[/tex]
= [tex]41 \ ppm[/tex]
The amount of free SO₂ to be added will be:
= [tex]41\times 225[/tex]
= [tex]9225 \ mg[/tex]
∵ 1000 mg = 1 g
So,
= [tex]9225\times \frac{1}{1000}[/tex]
= [tex]9.225[/tex]
Thus,
"9.225 g" should be added.
A student performs an experiment similar to Experiment 1 using hydrochloric acid (HCl) and potassium hydroxide (KOH). The mass of the hydrochloric acid solution is 250.000 g. After combining the HCl and KOH, the final combined mass is 400.000 g. Given what you have learned about the conservation of mass in this experiment, what must have been the mass of the KOH solution
Answer:
150.000 g
Explanation:
The law of conservation of mass states that the mass of reactants and products of a reaction must be equal to one another.
In other words, for this case:
Mass of KOH + Mass of HCl = Mass of ProductsWe are given all required data to calculate the mass of the KOH solution:
Mass of KOH + 250.000 g = 400.000 gMass of KOH = 150.000 gWhat happens when Sulphur dioxide (so2) gas is passed through an acidified solution of hydrogen . sulfide (H₂S) gas :
Answer:
When SO
2
is passed through an acidified solution of H
2
S, sulphur is precipitated out according to the reaction.
2H
2
S+SO
2
→2H
2
O+3S
Predict the products of below reaction, and whether the solution at equilibrium will be acidic, basic, or neutral.
N2O5 + 3H2O → __________
Answer: The product of the given reaction is [tex]HNO_{3}[/tex] and the solution at equilibrium will be acidic.
Explanation:
When two or more chemical substances react together then it forms new substances and these new substances are called products.
For example, [tex]3N_{2}O_{5} + 3H_{2}O \rightarrow 6HNO_{3}[/tex]
This shows that nitric acid [tex](HNO_{3})[/tex] is the product formed and it is an acidic substance.
Hence, the solution at equilibrium will be acidic in nature.
Thus, we can conclude that the product of the given reaction is [tex]HNO_{3}[/tex] and the solution at equilibrium will be acidic.
The doctor has ordered Claforan 1 g in 100 ml D5W to run IV piggyback for 30 minutes twice daily. The pharmacy sends Claforn 2 g in a powdered form, which when reconstituted has a concentration of 180 mg Claforan per ml. How much Claforn will you add to the bag of D5W
Answer:
0.111 g
Explanation:
1 g = 1000 mg
Doctor ordered the following concentration of Claforan:
C = 1 g/100 mL x 1000 mg/1 g = 10 mg/mL
If we add 2 g iof Claforan, we obtain:
2 g Claforn ---- 180 mg/mL Claforan
To reach a concentration equal to C (10 mg/mL), we need:
10 mg/mL Claforan x 2 g Claforn/(180 mg/mL Claforan) = 0.111 g Claforn
Therefore, we have to add 0,111 g (111 mg) of Claforn to the bag of 100 ml D5W to obtain the ordered concentration of 10 mg/mL Claforan.
Que es la actividad física y en qué mejora
Which of the following describes an organisms habitat?
A) where the organism lives
B) how the organism moves
C) what the organism eats
D) what eats the organism
Answer:
A) habitat
Explanation:
a habitat is essentially the organisms "home". also known as a "niche"
A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature changes from 5° C to 30° C? (Neglect the expansion of the beaker, evaporation of alcohol and absorption of water vapor by alcohol.) The volume coefficient of expansion γγ for alcohol = 1.12 x 10-4 K-1
Answer:
"1.4 mL" is the appropriate solution.
Explanation:
According to the question,
[tex]v_0=500[/tex][tex]\alpha =1.12\times 10^{-4}[/tex][tex]\Delta \epsilon = 25[/tex]Now,
Increase in volume will be:
⇒ [tex]\Delta V = \alpha\times v_0\times \Delta \epsilon[/tex]
By putting the given values, we get
[tex]=1.12\times 10^{-4}\times 500\times 25[/tex]
[tex]=1.12\times 10^{-4}\times 12500[/tex]
[tex]=1.4 \ mL[/tex]
In some sheep, the presence of horns is produced by an autosomal allele that is dominant in males and recessive in females.A horned female is crossed with a hornless male. One of the resulting F1 females is crossed with a hornless male. What proportion of the male and female progeny from this cross will have horns?(5 marks)
Answer:
1/2 f1 will cross
Explanation:
answer it
In a single displacement reaction Zinc can displace ALL but…
Iron
Nickel
Calcium
Lead
Answer:
Calcium
Explanation:
Zinc cannot displace Ca because calcium is above it in the reactivity series
Which one of the following molecule is planer?
a. NF3 c. PH3
b. BH3 d. NCl3
Answer:
option a
hope helps you
have a great day
Consider the titration of 30 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: a) 0 mL; b) 10 mL; c) 20 mL; d)35 mL; e) 36 mL; f) 37 mL.
Answer:
a)10.87
b)9.66
c)9.15
d)7.71
e) 5.56
f) 3.43
Explanation:
tep 1: Data given
Volume of 0.030 M NH3 solution = 30 mL = 0.030 L
Molarity of the HCl solution = 0.025 M
Step 2: Adding 0 mL of HCl
The reaction: NH3 + H2O ⇔ NH4+ + OH-
The initial concentration:
[NH3] = 0.030M [NH4+] = 0M [OH-] = OM
The concentration at the equilibrium:
[NH3] = 0.030 - XM
[NH4+] = [OH-] = XM
Kb = ([NH4+][OH-])/[NH3]
1.8*10^-5 = x² / 0.030-x
1.8*10^-5 = x² / 0.030
x = 7.35 * 10^-4 = [OH-]
pOH = -log [7.35 * 10^-4]
pOH = 3.13
pH = 14-3.13 = 10.87
Step 3: After adding 10 mL of HCl
The reaction:
NH3 + HCl ⇔ NH4+ + Cl-
NH3 + H3O+ ⇔ NH4+ + H2O
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.010 L = 0.00025 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.00025 =0.00065 moles
Moles HCl = 0
Moles NH4+ = 0.00025 moles
Concentration at the equilibrium:
[NH3]= 0.00065 moles / 0.040 L = 0.01625M
[NH4+] = 0.00625 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.00625/0.01625)
pOH = 4.34
pH = 9.66
Step 3: Adding 20 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.020 L = 0.00050 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.00050 =0.00040 moles
Moles HCl = 0
Moles NH4+ = 0.00050 moles
Concentration at the equilibrium:
[NH3]= 0.00040 moles / 0.050 L = 0.008M
[NH4+] = 0.01 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.01/0.008)
pOH = 4.85
pH = 14 - 4.85 = 9.15
Step 4: Adding 35 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.035 L = 0.000875 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.000875 =0.000025 moles
Moles HCl = 0
Moles NH4+ = 0.000875 moles
Concentration at the equilibrium:
[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M
[NH4+] = 0.000875 M / 0.065 L = 0.0135 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.0135/3.85*10^-4)
pOH = 6.29
pH = 14 - 6.29 = 7.71
Step 5: adding 36 mL HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.036 L = 0.0009 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.0009 =0 moles
Moles HCl = 0
Moles NH4+ = 0.0009 moles
[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M
Kw = Ka * Kb
Ka = 10^-14 / 1.8*10^-5
Ka = 5.6 * 10^-10
Ka = [NH3][H3O+] / [NH4+]
Ka =5.6 * 10^-10 = x² / 0.0136
x = 2.76 * 10^-6 = [H3O+]
pH = -log(2.76 * 10^-6)
pH = 5.56
Step 6: Adding 37 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.037 L = 0.000925 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.000925 =0 moles
Moles HCl = 0.000025 moles
Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M
pH = -log 3.73*10^-4= 3.43
The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:
a) pH = 10.86
b) pH = 9.66
c) pH = 9.15
d) pH = 7.70
e) pH = 5.56
f) pH = 3.43
Calculating the pH a) 0 mL
Initially, the pH of the solution is given by the dissociation of NH₃ in water.
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (1)
The constant of the above reaction is:
[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex] (2)
At the equilibrium, we have:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (3)
0.030 M - x x x
[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]
After solving for x and taking the positive value:
x = 7.18x10⁻⁴ = [OH⁻]
Now, we can calculate the pH of the solution as follows:
[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]
Hence, the initial pH is 10.86.
b) 10 mL
After the addition of HCl, the following reaction takes place:
NH₃ + HCl ⇄ NH₄⁺ + Cl⁻ (4)
We can calculate the pH of the solution from the equilibrium reaction (3).
[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)
Finding the number of moles of NH₃ and NH₄⁺
The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:
[tex] n_{b} = n_{i} - n_{HCl} [/tex] (6)
[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]
[tex] n_{a} = n_{HCl} [/tex] (7)
[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]
Calculating the concentrations of NH₃ and NH₄⁺The concentrations are given by:
[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex] (8)
[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex] (9)
Calculating the pHAfter entering the values of Ca and Cb into equation (5) and solving for x, we have:
[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]
x = 4.54x10⁻⁵ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]
Hence, the pH is 9.66.
c) 20 mLWe can find the pH of the solution from the reaction of equilibrium (3).
Calculating the concentrations of NH₃ and NH₄⁺The concentrations are (eq 8 and 9):
[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]
[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]
Calculating the pHAfter solving the equation (5) for x, we have:
[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]
x = 1.40x10⁻⁵ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]
So, the pH is 9.15.
d) 35 mLWe can find the pH of the solution from reaction (3).
Calculating the concentrations of NH₃ and NH₄⁺[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]
[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]
Calculating the pHAfter solving the equation (5) for x, we have:
[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]
x = 5.013x10⁻⁷ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]
So, the pH is 7.70.
e) 36 mL Finding the number of moles of NH₃ and NH₄⁺[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]
[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]
Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.
At the equilibrium, we have:
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺
Ca - x x x
[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]
[tex] Ka(Ca - x) - x^{2} = 0 [/tex] (10)
Calculating the acid constant of NH₄⁺
We can find the acid constant as follows:
[tex] Kw = Ka*Kb [/tex]
Where Kw is the constant of water = 10⁻¹⁴
[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]
Calculating the pH
The concentration of NH₄⁺ is:
[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]
After solving the equation (10) for x, we have:
x = 2.78x10⁻⁶ = [H₃O⁺]
Then, the pH is:
[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]
Hence, the pH is 5.56.
f) 37 mLNow, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).
Calculating the concentration of HCl
[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]
Calculating the pH
[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]
Therefore, the pH is 3.43.
Find more about pH here:
brainly.com/question/491373
I hope it helps you!
1. Which of the combinations in the lab activity had indications that a chemical change occured? Defend your argument with evidence.
2. Which of the combinations in the lab activity had indications that a physical change occured? Defend your argument with evidence.
3. Are all physical changes reversible? Explain your answer using an example you've observed in your everyday life.
4. Give an example of something you've observed in your everyday life that is a chemical reaction. How did you know it was a chemical reaction?
Answer:The green growing on the penny of copper and the rust forming on the nail of iron are chemical changes. Boiling away salt water, scraping iron filings from a mixture of sand with a magnet, and breaking a rock with a hammer, are physical changes.
Explanation:
PLEASE HELP!!
Solutions Pre-Lab Questions:
In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.
1. Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).
2. Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.
(Hint: Use molarity = to find the moles of drink mix, then convert moles to grams using a mole conversion.)
3. What mass of powdered drink mix is needed to make a 0.5 M solution of 100 mL?
Answer:
Explanation:
C₁₂H₂₂O₁₁
1 )
Molar mass = 12 x 12 + 22 x 1 + 11 x 16
= 144 + 22 + 176
= 342 g
2 )
100 mL of 1.0 M will contain 1.0 x0.100 = .1 mole of sucrose
0.1 mole of sucrose = 0.1 x 342 g = 34.2 g of sucrose.
So , mass of sucrose required is 34.2 g .
3 )
100 mL of .5 M sucrose = .100 x .5 mole of sucrose
= .05 mole of sucrose
.05 mole of sucrose = .05 x 342 g = 17.1 g of sucrose .
So , mass of sucrose required is 17.1 g .
Given 0.60 mol CO2, 0.30 mol CO, and 0.10 mol H20, what is the partial pressure of the CO if the total pressure of the mixture was 0.80 atm?
Answer:
Explanation:
/ means divided by
* means multiply
1. formula is
partial pressure = no of moles(gas 1)/ no of moles(total)
0.30 mol CO/0.60 mol CO2 + 0.30 mol CO + 0.10 mol H20 ->
.3/(.6+.3+.1) =
.3/1 =
.3 =
partial pressure of CO
2.
.3 * .8 atm = .24
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quizlet
The partial pressure of the CO is 0.24 atm if the total pressure of the mixture was 0.80 atm.
Dalton's Law of Partial pressureDalton's Law of partial pressure states that the total pressure exerted by non reacting gaseous mixture at a constant temperature and given volume is equal to the sum of partial pressure of all gases.
Dalton's Law of partial pressure using mole fraction of gas
Partial pressure of carbon monoxide (CO) = Mole fraction of carbon monoxide (CO) × Total pressure
Now, we have to find the first mole fraction of CO
Mole fraction of carbon monoxide (CO) = [tex]\frac{\text{moles of solute}}{\text{total moles of solute}}[/tex]
= [tex]\frac{\text{moles of CO}}{\text{moles of CO}_2 + \text{moles of CO} + \text{moles of H}_{2}O}[/tex]
= [tex]\frac{0.30}{0.60 + 0.30 + 0.10}[/tex]
= [tex]\frac{0.30}{1}[/tex]
= 0.3
Now, put the value in above equation, we get that
Partial pressure of carbon monoxide (CO)
= Mole fraction of carbon monoxide (CO) × Total pressure
= 0.3 × 0.8
= 0.24 atm
Thus, the partial pressure of the CO is 0.24 atm is the total pressure of the mixture was 0.80 atm.
Learn more about the Dalton's Law of partial Pressure here: https://brainly.com/question/14119417
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Enter a balanced equation for the reaction between solid nickel(II)(II) oxide and carbon monoxide gas that produces solid nickel and carbon dioxide gas. Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer: A balanced equation for the given reaction is [tex]NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)[/tex].
Explanation:
The reaction equation will be as follows.
[tex]NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)[/tex]
Number of atoms on the reactant side is as follows.
O = 2C = 1Number of atoms on the product side is as follows.
Ni = 1O = 2C = 1Since number of atoms on both the reactant and product sides are equal. Hence, the reaction equation is balanced.
Thus, we can conclude that a balanced equation for the given reaction is [tex]NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)[/tex].
Why ethanol is used in pectin extraction
Explanation:
is responsible for interrupting the interaction between pectins and solvent molecules
an emerald can be described as...
Answer:
green gemstone
Explanation:
hope this helps someone
Ethylene glycol flows at 0.01 kg/s through a 3-mm diameter, thin-walled tube. The tube is coiled and submerged in well-stirred water bath maintained at 25°C. If the fluid enters the tube at 85°C, what heat rate and tube length are required for the fluid to leave at 35°C?
Answer:
heat rate= 1281W
length = 15.8m
Explanation:
we have this data to answer this question with
Tmi = 85 degrees
Tmo = 35 degrees
Ts = 25 dgrees
flow rate = 25 degrees
using engine oil property from table a-5
Tm = Tmo - TMi/2 = 333k
u =0.522x10⁻²
k = 0.26
pr = 51.3
cp = 2562 J/kg.k
mcp(Tmo-Tmi) =
0.01 x 2562(35-85)
= 1281 W
we find the change in Tim
= [(35-25)-(85-25)]/ln[(35-25)/(85-25)]
= -50/ln0.167
= -50/-1.78976
= 27.9°c
we finf the required reynold number
4x0.01/πx0.003x0.522x10⁻²
= 0.04/0.00004921
= 812.8
= 813
we find approximate correlation
NuD = hd/k
NuD = 3.66
3.66 = 0.003D/0.26
cross multiply
0.003D = 3.66x0.26
D = 3.66x0.26/0.003
= 317.2
As = 1281/317x27.9
= 0.145
As = πDL
L = As/πD
= 0.145/π0.003
= 0.145/0.009429
L = 15.378
Sugar is added to water and initially completely dissolves, but eventually solid sugar collects on the bottom of the container. Sugar and water are ________partially miscible . This produces a dynamic equilibrium. Ethanol (a liquid) is added to water and only a single layer is observed no matter how much ethanol is added. Ethanol and water are__________
Answer: Sugar is added to water and initially completely dissolves, but eventually solid sugar collects on the bottom of the container. Sugar and water are both partially miscible. This produces a dynamic equilibrium. Ethanol (a liquid) is added to water and only a single layer is observed no matter how much ethanol is added. Ethanol and water are miscible.
Explanation:
When a substance (solute) dissolves partially in a solvent then it is known as partially miscible in the solvent. In such cases, a small amount of solute remains at the bottom of solution.
If a solute dissolves completely in solvent like water such that only one layer is seen in the solution then it means that the solute is miscible in solvent.
Thus, we can conclude that sugar is added to water and initially completely dissolves, but eventually solid sugar collects on the bottom of the container. Sugar and water are both partially miscible. This produces a dynamic equilibrium. Ethanol (a liquid) is added to water and only a single layer is observed no matter how much ethanol is added. Ethanol and water are miscible.
