Leo carefully pipets 50.0 mL of 0.500 M NaOH into a test tube. She places the test tube
into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.250 M HCl into
another test tube. When Leo reaches to put this test tube of acid into the beaker along
with test tube of base she accidentally knocks the test tubes together hard enough to
break them and their respective contents combine in the bottom of the beaker. Is the
solution formed from the contents of the two test tubes acidic or basic? What is the pH of
the resulting solution?
Please answer below questions one by one to assist you receive full credits
(Alternatively, you can discard my hints below, solve the problem using your own way
and send me the picture/copy of your complete work through email)
The mole of NaOH before mixing is
mol (save 3 significant figures)
The mole of HCl before mixing is
mol (save 4 significant figures)
After mixing, the solution is
(choose from acidic or basic)
The total volume of mixture is
L (save 3 significant figures)
The concentration of [OH-] is
M (save 3 significant figures)
The concentration of [H'l is
M (save 3 significant figures)
Let's consider the neutralization reaction between HCl and NaOH.
NaOH + HCl ⇒ NaCl + H₂O
To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.
NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol
HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol
Now, let's determine the reactant in excess and the remaining moles of that reactant.
NaOH + HCl ⇒ NaCl + H₂O
Initial 0.0200 0.0188
Reaction -0.0188 -0.0188
Final 1.20 × 10⁻³ 0
The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³ moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:
[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M
NaOH is a strong base according to the following equation.
NaOH ⇒ Na⁺ + OH⁻
The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.
The pOH is:
pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02
We will calculate the pH using the following expression.
pH = 14.00 - pOH = 14.00 - 2.02 = 11.98
The pH is 11.98. Since pH > 7, the solution is basic.
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9.7300x10^2 + 9.8700x10^3
? × 10^?
Answer:
19.6 x 10⁵
1.96 x 10⁶
I hope it's helps you
Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 55.8 g of hydrobromic acid is mixed with 17. g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Answer:
21.4g of HBr is the minimum mass that could be left over.
Explanation:
Based on the reaction:
HBr + NaOH → NaBr + H2O
1 mole of HBr reacts per mole of NaOH
To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:
Moles NaOH -40.0g/mol-
17g * (1mol/40.0g) = 0.425 moles NaOH
Moles HBr -Molar mass: 80.91g/mol-
55.8g * (1mol/80.91g) = 0.690 moles HBr
The difference in moles is:
0.690 moles - 0.425 moles =
0.265 moles of HBr could be left over
The mass is:
0.265 moles * (80.91g/mol) =
21.4g of HBr is the minimum mass that could be left over.According to the EPA Lead and Copper Rule (LCR), the action level for Pb in drinking water (the level at which threat to human health requires public notification and action towards mitigation) is 15 ppb. If you were to add enough phosphate to the system
saturated with respect to Pb3(PO4)2(s), would the [Pb2+] be below the action limit?
Answer:
The right answer is "105.17 ppb".
Explanation:
According to the question,
The amount of [tex]Pb^{2+}[/tex] in ppb will be:
= [tex]0.5076\times 10^{-6}\times 207.2\times 106[/tex]
= [tex]105.17 \ ppb[/tex]
Thus, the amount of [tex]Pb^{2+}[/tex] is above action limit.
what is a compound?And what are some common examples of a compound?
Answer:
Compounds are substances made from atoms of different elements joined by chemical bonds. Common examples are water (H2O), salt (sodium chloride, NaCl), methane (CH4).
If you drip an ink drop into a cup of water and wait for a few seconds, all the water will be colored with the ink. This experiment is an example of facilitated diffusion ?
true
false
Answer:
false, it is not an example of facilitated diffusion
Answer:
TrueExplanation:
When a drop of ink added into the water gradually moves in the whole quantity of water due to this entire water turns into blue color. This is nothing but the diffusion of ink particles into the water molecules. This is because water, as well as ink molecules, are in random motion due to the motion of ink substance.
Given the standard enthalpy changes for the following two reactions: (1) 2Fe(s) + O2(g)2FeO(s)...... ΔH° = -544.0 kJ (2) 2Zn(s) + O2(g)2ZnO(s)......ΔH° = -696.6 kJ what is the standard enthalpy change for the reaction:
Answer:
-76.3 kJ
Explanation:
Here is the complete question
Given the standard enthalpy changes for the following two reactions:
(1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ
(2) 2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:
(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)......ΔH° = ?
Solution
Since (1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ
reversing the reaction, we have
2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)
Adding reactions (2) and (3), we have
2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)
2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ (2)
This gives
2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)......ΔH° =
The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)
= +544.0 kJ - 696.6 kJ)
= -152.6 kJ
Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)
we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).
So, ΔH° = -152.6 kJ/2 = -76.3 kJ
So, the standard enthalpy change for the reaction
FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ
Consider the following titration for these three questions:
1.00 L of 2.00 M HCl is titrated with 2.00 M NaOH.
a. How many moles of acid are equal to one equivalent in this titration?
b. How many moles of HCl are found in solution at the halfway point of the titration?
c. How many liters of base will be needed to reach the equivalence point of the titration?
Answer:
a. 1 mole of acid is equal to one equivalent.
b. 1.00 moles of HCl are found.
c. 1L of 2.00M NaOH is needed to reach the equivalence point
Explanation:
HCl reacts with NaOH as follows:
HCl + NaOH → NaCl + H2O
Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1
a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent
b. The initial moles of HCl are:
1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl
At the halfway point, the moles of HCl are the half, that is:
1.00 moles of HCl are found
c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:
2.00moles NaOH * (1L / 2.00mol) =
1L of 2.00M NaOH is needed to reach the equivalence point
Match each land resource to its use.
clay - used to make steel
iron ore - used to make batteries
salt - used to make pottery and tiles
aggregate - used in construction
graphite - used as a flavoring in food
i will give 10 points and brainliest!!!
Answer:
answer in picture
Explanation:
Two methods by which we can conserve water and water the plants.
Answer:
Two methods which help us to conserve water are:
Sprinkler irrigation system: this irrigation has an arrangement of vertical pipes with rotating nozzles on the top. It is more useful in the uneven and sandy land where sufficient water is not available.
Drip irrigation system: this irrigation system has an arrangement of pipes or tubes with very small holes in them to water plants drop by drop just at the base of the root. It is very efficient as water is not wasted at all.
Consider the preparation of methyl benzoate by reacting benzoic acid with methanol using sulfuric acid as a catalyst. Reaction scheme of benzoic acid with methanol, conc. sulfuric acid, and heat over the arrow, and methyl benzoate and water as products. Calculate the molar masses of the reactant and product. Report molar masses to 1 decimal place. Molar mass of benzoic acid g/mol Molar mass of methyl benzoate
Answer:
See explanation
Explanation:
The molecular mass is the sum of the relative atomic masses of all the atoms in the molecule.
The relative atomic mass of reactants and products are calculated as follows;
Benzoic acid is C7H6O2 hence the molar mass of benzoic acid is ;
7(12) + 6(1) + 2(16) = 84 + 6 + 32 = 122.0 g/mol
Methyl benzoate is C8H8O2
8(12) + 8(1) + 2(16) = 96 + 8 + 32 = 136.0 g/mol
Rank the compounds NH3, CH4, and PH3 in order of decreasing boiling point. Choices: A) NH3 > CH4 > PH3 B) CH4 > NH3 > PH3 C) NH3 > PH3 > CH4 D) CH4 > PH3 > NH3 E) PH3 > NH3 > CH4
Answer:
C) NH3 > PH3 > CH4
Explanation:
The boiling point of a substance depends on the nature of intermolecular interaction between the molecules of the substance. The greater the magnitude of intermolecular interaction between the molecules of the substance, the higher the boiling point of the substance.
Both NH3 and PH3 have intermolecular hydrogen bonding between their molecules. However, since nitrogen is more electronegative than phosphorus, the magnitude of intermolecular hydrogen bonding in NH3 is greater than in PH3 hence NH3 has a higher boiling point than PH3.
CH4 molecules only have weak dispersion forces between them hence they exhibit the lowest boiling point.
Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A Write chemical equations for first ionization step of carbonic acid. Express your answer as a chemical equation. Identify all of the phases in your answer.
Solution :
[tex]H_2CO_3[/tex] is considered a diprotic acid.
Sp it can dissociate in solution by giving two protons.
Chemical equations for the first step of carbonic acid is :
First ionization
[tex]$H_2CO_3(aq) + H_0(1) \rightleftharpoons H_.O^+(aq) + HCO_3^-(aq)$[/tex]
Equilibrium constant expression is
[tex]$K_{a}_{1}=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$[/tex]
Second ionization -
[tex]$HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CO_3^{2-}(aq)$[/tex]
Equilibrium constant expression is
[tex]$K_{a2}=\frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]}$[/tex]
write anode and cathode in Zn-Ag galvanic cell
Explanation:
Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).
By convention in standard cell notation, the anode is written on the left and the cathode is written on the right. So, in this cell: Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).
A chemist is preparing to carry out a reaction that requires 5.75 moles of hydrogen gas. The chemist pumps the hydrogen into a 10.5 L rigid steel container at 20.0 °C. To what pressure, in kPa, must the hydrogen be compressed? (Show all work for full credit and circle your final answer) *
Answer:
The hydrogen must be compressed to 1333.13302 kPa.
Explanation:
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case:
P= ?V= 10.5 Ln= 5.75 molesR= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 20 C= 293 K (being 0 C= 273 K)Replacing:
P* 10.5 L= 5.75 moles* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] * 293 K
Solving:
[tex]P=\frac{5.75 moles* 0.082 \frac{atm*L}{mol*K} * 293 K}{10.5 L}[/tex]
P= 13.157 atm
If 1 atm is equal to 101.325 kPa, then 13.157 atm is equal to 1333.13302 kPa.
The hydrogen must be compressed to 1333.13302 kPa.
Ortho and para hydrogen are....... a). molecular form. b). Nuclear form. c) allotropic form. d). All
Ortho and para hydrogen are nuclei forms
Which of the following compounds would you expect to be an electrolyte?
N2
CH4
H2O
O2
КСІ
Answer:
N2 but i really didn't know
The compound that would be expected to be an electrolyte is : ( A ) N₂
What is an electrolyte
An electrolyte is any subsatnce which conducts electircity when dissolved in a solvent such as water. From the question the compound that can conduct electricty when dissolved in water is N₂
Hence we can conclude that The compound that would be expected to be an electrolyte is : ( A ) N₂
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#SPJ2
A buffer is a solution that: Select one: a. Results from mixing a strong acid and a strong base. b. When added to another solution, decreases the pH. c. When added to another solution, increases the pH. d. Prevents a drastic change in pH when an acid or base enters a solution.
Answer:
d. Prevents a drastic change in pH when an acid or base enters a solution.
Explanation:
The purpose of a buffer is to resist pH change and keep the solution relatively stable.
A fusion reaction releases energy because the binding energy of the resulting nucleus:______.
a. is released in the process.
b. is equal to the binding energy of the original nuclei.
c. is absorbed in the process.
d. is less than the binding energy of the original nuclei.
e. is greater than the binding energy of the original nuclei.
Answer:
a. is released in the process
Explanation:
In fusion reaction the nucleus is unstable so it releases its binding energy resulting in decreasing its mass so it becomes more stable.
Is Water and kerosine a mixture
Answer:
No.Kerosene oil and water do not mix with each other and form two separate layers.
Answer:
No
Explanation:
They cannot be mixed together they will form upper and lower layer
The mole fraction of NaCl in an
aqueous solution is 0.132. How
many moles of NaCl are present in
1 mole of this solution?
Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol
Answer:
Moles of water are 0.868
Explanation:
Question 14
2 pts
A chemist wants to make 100 mL of a 0.500 M solution of NaCl. They have a
stock solution of 1.2 M NaCl. How much of the original stock solution do they
need to make their new dilute solution?
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 1.2 M
Molarity of diluted solution (M₂) = 0.5 M
Volume of diluted solution (V₂) = 100 mL
Volume of stock solution needed (V₁) =?The volume of stock solution needed can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂1.2 × V₁ = 0.5 × 100
1.2 × V₁ = 50
Divide both side by 1.2
V₁ = 50 / 1.2
V₁ ≈ 42 mLThus, 42 mL of the stock solution is needed.
Learn more: https://brainly.com/question/24219233
Answer:
They need 41.7 mL of the original stock solution.
Explanation:
We can use the following equation for dilutions:
Cc x Vc = Cd x Vd
Where Cc and Vc are the concentration and volume values in the concentrated condition, whereas Cd and Vd are the concentration and volume values in the diluted condition.
The concentrated solution is the original stock solution, and it has:
Cc = 1.2 M
The diluted solution must be:
Cd = 0.500 M
Vd = 100 mL
So, we have to calculate Vc. For this, we replace the data in the equation:
[tex]V_{c} = \frac{C_{d} V_{d} }{C_{c} } = \frac{(0.500 M)(100 mL)}{1.2 M} = 41.7 mL[/tex]
Therefore, 41.7 mL of 1.2 M original stock solution are required to make 100 mL of a diluted solution with a concentration of 0.500 M.
Write the chemical formula for the following:
a. The conjugate acid of amide ion, NH₂-
b. The conjugate base of nitric acid, HNO₃
Follow the rules of Bronsted Lowry theory
Acids take a HBases donate a HSo
#a
NH_2-
Add a H
Conjugate acid is NH_3#b
HnO_3
Take a H
Conjugate base is NO_3-#1
Conjugate acid means one H+ is added
NH_2+H+=NH_3sw
#2
Conjugate base means donate a H+
HNO_3-H=NO_3-Hydrogen chloride decomposes to form hydrogen and chlorine, like this:
2HCl(g) + H2(g) â Cl2(g)
Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen chloride, hydrogen, and chlorine has the following composition:
compound pressure at equilibrium
HCl 84.4 atm
H2 77.9 atm
Cl2 54.4
Required:
Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.
Solution :
Given :
Partial pressure of HCl, [tex]$P_{HCl}$[/tex] = 84.4 atm
Partial pressure of [tex]H_2[/tex], [tex]$P_{H_2}$[/tex] = 77.9 atm
Partial pressure of [tex]Cl_2[/tex], [tex]$P_{Cl_2}$[/tex] = 54.4 atm
Reaction :
[tex]$2HCl (g) \leftrightharpoons H_2(g) + Cl_2(g)$[/tex]
Using equilibrium concept,
[tex]$k_p=\frac{(P_{H_2})(P_{Cl_{2}})}{(P_{HCl})^2}$[/tex]
[tex]$k_p=\frac{77.9 \times 54.4}{(84.4)^2}$[/tex]
[tex]$k_p=0.594$[/tex]
[tex]k_p=0.59[/tex] (in 2 significant figures)
or [tex]k_p=5.9 \times 10^{-1}[/tex]
how many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3 ?
Answer:
Explanation:
The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.
0.2175,0.0725 moles of [tex]H_2[/tex] and [tex]N_2[/tex] can be formed by the decomposition of 0.145 mol of ammonia, [tex]NH_3[/tex]
The reaction for the decomposition of ammonia is as follows:-
[tex]N_2+3H_2 \rightarrow 2NH_3[/tex]
Calculate the mole of [tex]H_2[/tex] and [tex]N_2[/tex] as follows:-
[tex]Mole\ of \ H_2=0.145\ mol\ NH_3\times\frac{3\ mol\ H_2}{2\ mol\ NH_3} \\\\=0.2175\ mol\ H_2[/tex]
[tex]Mole\ of \ N_2=0.145\ mol\ NH_3\times\frac{1\ mol\ N_2}{2\ mol\ NH_3} \\\\=0.0725\ mol\ H_2[/tex]
Hence, the number of moles of [tex]H_2[/tex] and [tex]N_2[/tex] are 0.2175 mol, and 0.0725 mol.
To know more about:-
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Assume that all products containing “Bromide” or an ingredient ending in “-ol” are toxic.
Product B and E are toxic because they contain "BROMIDE" or ingredients that end in 'ol'
Based on the directions given in the information of this question, any product containing "bromide" or containing an ingredient that ends in "ol" is assumed to be TOXIC.
After carefully evaluating the ingredient contents of each product in the image attached to this question, it was realized that product B contains "pyridostigmine bromide" as an ingredient while product E contains "butorphanol" as an ingredient. Hence, in accordance to the guide given in this question, products B and E are toxic.
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Rubric #2
Forensic Science
1. Define Nucleus.
2. Define Cytoplasm.
3. Define Cell Membrane.
4. Define DNA.
5. Define Plant.
6. Define Chlorophyll.
7. Define Photosynthesis.
8. How do Plant cells and Animal cells differ?
9. Define Cell Wall.
10. Define Vacuole.
11. Why do cells differentiate in multicellular organisms?
12. Define Multicellular.
13. Complete the Eukaryotic cells and Cell Differentiation assessment.
https://clever.discoveryeducation.com/learn/techbook/units/95c20a43-6d3d-40d3-
848d-89929101140d/concepts/co0fef01-33e7-4116-8819.
143e289e15ba/tabs/6e1551ab-57b8-42d4-8e5b-25549791c760/pages/de4182af-aa 60-
454f-ae5e-28df6f4eb3ac
Explanation:
1. Nucleus is a memberane bound organelle that contains cell,s chromosomes.
When a marble is dropped into a beamer of water
Answer:
The water will rise.Explanation:
hope this helps you
-Sweety<3The mass of the marble is greater than that of the water. The marble weighs more than an equivalent volume of the water. The force from dropping the marble breaks the surface tension of the water. The marble has greater mass and volume than the water.
Calculate the volume of a 89.51 g sample of carbon dioxide at 281.8 K and 843.9 torr. Round your answer
to the nearest L. Do not include units.
Answer:
1,000.000
Explanation:
Classify the following oxides as acidic, basic, amphoteric, or neutral:
(a) Select the acidic oxides:
A. CO₂
B. N₂O₅
C. Al₂O₃
D. NO
E. SO₃
(b) Select the basic oxides:
A. CaO
B. CO
C. SO₃
D. K₂O
E. BaO
(c ) Select the amphoteric oxides:
A. K₂O
B. Al₂O₃
C. CaO
D. CO₂
E. SnO₂
d) Select the neutral oxides:
A. CO
B. NO
C. SNO₂
D. N₂O₅
E. BaO
Answer:
(a).
» E. SO₃, sulphur trioxide.
(b).
» A. CaO, Calcium oxide.
» D. K₂O, potassium oxide.
» E. BaO, barium oxide.
(c).
» B. Al₂O₃, Aluminium oxide.
» E. SnO₂, tin (IV) oxide.
(d).
» A. CO, carbon monoxide.
» B. NO, nitrogen monoxide.
Explanation:
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