Explanation:
To calculate [H3O+] in the solution we must first find the pH from the [ OH-]
That's
pH + pOH = 14
pH = 14 - pOH
To calculate the pOH we use the formula
pOH = - log [OH-]
And [OH-] = 5.5 × 10^-5 M
So we have
pOH = - log 5.5 × 10^ - 5
pOH = 4.26
Since we've found the pOH we can now find the pH
That's
pH = 14 - 4.26
pH = 9.74
Now we can find the concentration of H3O+ in the solution using the formula
pH = - log H3O+
9.74 = - log H3O+
Find the antilog of both sides
H3O+ = 1.8 × 10^ - 10 MThe solution is basic since it's pH lies in the basic region.
Hope this helps you
1500 L has how many significants figures
Answer:
It has 2
Explanation:
The significant figures are 1 and 5!
Hope this helps:)
Methanol liquid burns readily in air. One way to represent this equilibrium is: 2 CO2(g) + 4 H2O(g)2 CH3OH(l) + 3 O2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) K1 = 2) CO2(g) + 2 H2O(g) CH3OH(l) + 3/2 O2(g) K2 = 3) 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)
Answer:
Answers are in the explanation
Explanation:
It is possible to obtain K of equilibrium of related reactions knowing the laws:
A + B ⇄ C K₁
C ⇄ A + B K = 1 /K₁
The inverse reaction has the inverse K equilibrium
2A + 2B ⇄ 2C K = K₁²
The multiplication of the coefficients of reaction produce a k powered to the number you are multiplying the coefficients
For the reaction:
2 CO2(g) + 4 H2O(g) ⇄ 2 CH3OH(l) + 3 O2(g) K
1) CH3OH(l) + 3/2 O2(g) ⇄ CO2(g) + 2 H2O(g)
This is the inverse reaction but also the coefficients are dividing in the half, that means:
[tex]K_1 = \frac{1}{k^{1/2}} = (1/K)^{1/2}[/tex]
2) CO2(g) + 2 H2O(g) ⇄ CH3OH(l) + 3/2 O2(g)
Here,the only change is the coefficients are the half of the original reaction:
[tex]K_2 = K^{1/2}[/tex]
3) 2CH3OH(l) + 3 O2(g) ⇄ 2 CO2(g) + 4 H2O(g)
This is the inverse reaction. Thus, you have the inverse K of equilibrium:
[tex]K_3 = \frac{1}{K}[/tex]
Aqueous ammonia is added to a mixture of silver chloride and water. Given that Kf for the reaction between Ag+ and NH3 is large, which of the following are true?
A) The free ions are favored over the complex ion.
B) The complex ion is favored over solid silver chloride.
C) The free Ag+ ion is unstable.
D) More silver chloride will precipitate.
Answer:
B) The complex ion is favored over solid silver chloride
C) The free Ag+ ion is unstable.
Explanation:
Hello,
In this case, since the dissociation of solid silver chloride occurs at equilibrium with a neglectable solubility product (very small Ksp), which means that the solid tends to remain undissolved:
[tex]AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]
By the addition of ammonia, the following reaction is favored:
[tex]Ag^+(aq)+2NH_3(aq)\rightleftharpoons [Ag(NH_3)_2]^+(aq)[/tex]
Which has a large equilibrium constant, which means that the formation of the complex is assured. In such a way, by addition of more ammonia, more complex will be formed, therefore B) The complex ion is favored over solid silver chloride is true. Moreover, C) The free Ag+ ion is unstable, since they tend to form the complex once they are formed by the solid silver chloride so it readily reacts.
Best regards.
A compound is found to contain 30.45 % nitrogen and 69.55 % oxygen by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 46.01 g/mol. What is the molecular formula for this compound?
Answer:
Empirical formulae is NO2
Molecular Formulae is NO2
Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
Find the volume if the density is 2.6 g/mL and the mass is 9.7 g.(5 pts)
Find the mass if the density is 1.6 g/cm3 and the volume is 4.1 cm3 (5 pts)
Find the density if the initial volume of water is 12.8 mL, the final volume is 24.6 mL and the mass of the object is 4.3 g. Make a drawing to show the water displacement using a graduated cylinder. (gdoc, gdraw)
Answer:
[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]
Explanation:
1. Density from mass and volume
[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]
2. Volume from density and mass
[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]
3. Mass from density and volume
[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]
4. Density by displacement
Volume of water + object = 24.6 mL
Volume of water = 12.8 mL
Volume of object = 11.8 mL
[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]
Your drawing showing water displacement using a graduated cylinder should resemble the figure below.
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppose 0.963 g of methane is mixed with 7.5 g of oxygen. Calculate the minimum mass of methane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
[tex]left\ over=0g[/tex]
Regards.
What compound is formed when methyloxirane (1,2-epoxypropane) is reacted with ethylmagnesium bromide followed by treatment with aqueous acid
Answer:
Pentan-2-ol
Explanation:
On this reaction, we have a Grignard reagent (ethylmagnesium bromide), therefore we will have the production of a carbanion (step 1). Then this carbanion can attack the least substituted carbon in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the treatment with aqueous acid, when we add acid the hydronium ion ([tex]H^+[/tex]) would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be attacked by the negative charge produced in the second step to produce the final molecule: "Pentan-2-ol".
See figure 1
I hope it helps!
Which of the following contains a nonpolar covalent bond?
O A. Co
B. NaCl
O C. 02
O D. HE
Answer:
The answer is o2
Explanation:
I took the test
How many atoms of oxygen are in one molecule of water (H2O)? one two four three
Answer:
there is one atom of oxygen and two atoms of hydrogen
Explanation:
Which of the following compounds is more soluble in a 0.10 M NaCN solution than in pure neutral water? Ca3(PO4)2 AgBr CaCO3 Mg(OH)2 NH4ClO4
Answer:
AgBr
Explanation:
Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.
However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.
The equation for the dissolution of AgBr in cyanide is shown below;
AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)
The density of concentrated nitric acid (HNO3) is 1.413 g/mL. What volume in liters would be occupied by a mass of 47.2 g?
Answer:
The volume that a mass of 47.2 g would occupy is 0.0334 L
Explanation:
Density is the property that matter, whether solid, liquid or gas, has to compress into a given space. Density is defined as the amount of mass it has per unit volume, that is, the ratio between the mass of a body and the volume it occupies:
[tex]Density=\frac{mass}{volume}[/tex]
This indicates that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.
In this case:
density= 1.413 g/mLmass= 47.2 gvolume=?Replacing:
[tex]1.413 \frac{g}{mL}=\frac{47.2 g}{volume}[/tex]
Solving:
[tex]volume=\frac{47.2 g}{1.413\frac{g}{mL} }[/tex]
volume=33.40 mL
Being 1,000 mL= 1 L:
volume= 0.0334 L
The volume that a mass of 47.2 g would occupy is 0.0334 L
Calculate LaTeX: \DeltaΔGº for a voltaic cell with Eºcell = +0.24 V if the overall reaction involves a 3 electron reduction.
Answer:
-69 kJ
Explanation:
Step 1: Given data
Standard cell potential (E°cell): +0.24 V
Electrons involved (n): 3 mol
Step 2: Calculate the standard Gibbs free energy change (ΔG°) for the voltaic cell
We will use the following expression.
ΔG° = -n × F × E°cell
where,
F is Faraday's constant (96,485 C/mol e⁻)
ΔG° = -n × F × E°cell
ΔG° = -3 mol e⁻ × 96,485 C/mol e⁻ × 0.24 V
ΔG° = -69 kJ
How many kg/hr of steam are produced by a 50HP boiler?
Answer:
Explanation:
50 HP = 50 x 746 watt
= 37300 watt
= 37300 J /s
heat produced in one hour = 60 x 60 x 37300 J
= 134280 x 10³ J
latent heat of vaporization = 2260 x 10³ J / kg .
for evaporation of water of mass 1 kg , 2260 x 10³ J of heat is required .
kg of water being evaporated by boiler per hour
= 134280 x 10³ / 2260 x 10³
= 59.41 kg
rate of production of steam
= 59.41 kg / hr .
In a buffer solution made of acetic acid and sodium acetate, if a small amount of acid is added, the added acid will react with whome?
Answer:
The acid reacts with the conjugate base producing more weak acid.
Explanation:
A buffer solution is defined as the mixture of a weak acid and its conjugate base or a weak base with its conjugate acid.
The acetic buffer, CH₃COOH/CH₃COO⁻, is in equilibrium with water as follows:
CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺
When an acid HX (Source of H₃O⁺) is added to the buffer, the reaction that occurs is:
CH₃COO⁻ + HX → CH₃COOH
The acid reacts with the conjugate base producing more weak acid.In fact, this is the principle of the buffer:
An acid reacts with the conjugate base producing weak acid. And the weak acid reacts with a base producing conjugate base
If a small amount of acid is added to an acetic acid-sodium acetate buffer, the acid will react with the acetate ion from sodium acetate.
We have a buffer formed by acetic acid and sodium acetate.
What is a buffer?A buffer is a solution used to resist abrupt changes in pH when an acid or a base is added.
How are buffers formed?They can be formed in 1 of 2 ways:
By a weak acid and its conjugate base.By a weak base and its conjugate acid.Our buffer is formed by a weak acid (acetic acid) and its conjugate base (acetate ion from sodium acetate).
When an acid (HX) is added, it is neutralized by the basic component of the buffer. The generic net ionic equation is:
H⁺ + CH₃COO⁻ ⇄ CH₃COOH
If a small amount of acid is added to an acetic acid-sodium acetate buffer, the acid will react with the acetate ion from sodium acetate.
Learn more about buffers here: https://brainly.com/question/24188850
Arrange the following in order of increasing boiling point: CH4, CH3CH3, CH3CH2Cl, CH3CH2OH. Rank from lowest to highest. To rank items as equivalent, overlap them.
Answer:
In order from lowest to highest:
Methane < Ethane < Chloroethene < Methanol
i.e: CH4 < CH3CH3 < CH3CH2OH < CH3CH2Cl
Explanation:
Compounds with stronger molecular fore have higher boiling points, thus making the molecules more difficult to pull apart. The presence of chains also increases the molecular dispersion. The dipole force of ethanol makes it have a very high boiling point.
I'm positive this explanation would suffice. Best of luck.
The order of increasing boiling points of the substances listed is; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH.
Intermolecular interactions occur between molecules. The boiling point and melting points of substances depends on the nature and magnitude of intermolecular interaction between the molecules of the substance.
The order of increasing boiling points of the substances listed is as follows; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH. CH3CH2OH has the highest boiling point due to intermolecular hydrogen bonds in the molecule. Though CH4 and CH3CH3 are both alkanes, CH3CH3 has a higher molecular mass and consequently greater dispersion forces and a higher boiling point.
Learn more: https://brainly.com/question/865531
Need help with chemistry questions
Answer:
1. oxidation
2. reduction
3. oxidation
4. oxidation
Explanation:
Oxidation and Reduction in terms of hydrogen
Oxidation and Reduction with respect to Hydrogen Transfer. Oxidation is the loss of hydrogen. Reduction is the gain of hydrogen.Oxidation and Reduction in terms of Oxygen
Oxidation and Reduction with respect to Oxygen Transfer. Oxidation is the gain of Oxygen. Reduction is the loss of Oxygen.The compound sodium hydroxide is a strong electrolyte. Write the transformation that occurs when solid sodium hydroxide dissolves in water. Use the pull-down boxes to specify states such as (aq) or (s).
Answer:
Solid sodium hydroxide dissolves in water to form an aqueous solution of ions.
NaOH(s) ⇌ Na+(aq) + OH–(aq) ΔH1 = ?
Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride.
NaOH(s) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH2 = ?
Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride.
Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH3 = ?
The electrolysis of molten AlCl 3 for 2.50 hr with an electrical current of 15.0 A produces ________ g of aluminum metal.
For each of the following reactions calculate the mass (in grams) of both the reactants that are required to form 15.39g of the following products.
a. 2K(s) + Cl2(g) → 2Cl(aq)
b. 4Cr(s) + 302(g) → 2Cr2O3(s)
c. 35r(s) + N2(g) → SraNa(s)
Answer:
a.
[tex]m_K=8.056gK\\ \\m_{Cl_2}=4.028gCl_2[/tex]
b.
[tex]m_{Cr}=10.51gCr\\ \\m_{O_2}=4.851gO_2[/tex]
c.
[tex]m_{Sr}=13.88gSr\\\\m_{N_2}=1.479gN_2[/tex]
Explanation:
Hello,
In this case, we proceed via stoichiometry in order to compute the masses of all the reactants as shown below:
a. [tex]2K+Cl_2\rightarrow 2KCl[/tex]
[tex]m_K=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{2molK}{2molKCl}* \frac{39.1gK}{1molK}=8.056gK\\ \\m_{Cl_2}=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{1molCl_2}{2molKCl}* \frac{70.9gCl_2}{1molCl_2}=4.028gCl_2[/tex]
b. [tex]4Cr+ 3O_2\rightarrow 2Cr_2O_3[/tex]
[tex]m_{Cr}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{4molCr}{2molCr_2O_3}* \frac{52gCr}{1molCr_2O_3}=10.51gCr\\ \\m_{O_2}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{3molO_2}{2molCr_2O_3}* \frac{32gO_2}{1molCr_2O_3}=4.851gO_2[/tex]
c. [tex]3Sr(s) + N_2(g) \rightarrow Sr_3N_2[/tex]
[tex]m_{Sr}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{3molSr}{1molSr_3N_2}* \frac{87.62gSr}{1molSr}=13.88gSr\\\\m_{N_2}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{1molN_2}{1molSr_3N_2}* \frac{28gN_2}{1molN_2}=1.479gN_2[/tex]
Regards.
19. For the following unbalanced equation: S (s) O2 (g) H2O (l) ----> H2SO4 (aq) At what temperature (K) does O2 have to be if the volume of the gas is 5.01 L with a pressure of 0.860 atm to produce 17.55 grams of H2SO4
Answer:
195.5 K
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2S + 3O2 + 2H2O → 2H2SO4
Next, we shall determine the number of mole in 17.55 g of H2SO4. This can be obtained as follow:
Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 + 64 = 98 g/mol
Mass of H2SO4 = 17.55 g
Mole of H2SO4 =...?
Mole = mass /Molar mass
Mole of H2SO4 = 17.55/98
Mole of H2SO4 = 0.179 mole.
Next, we shall determine the number of mole of O2 needed for the reaction. This can be obtained as illustrated below:
From the balanced equation above,
3 moles of O2 reacted to produce 2 moles of H2SO4.
Therefore, Xmol of O2 will react to produce 0.179 moleof H2SO4 i.e
Xmol of O2 = (3 x 0.179) / 2
Xmol of O2 = 0.2685 mole
Therefore, 0.2685 mole of O2 is needed for the reaction.
Finally, we shall determine the temperature of O2. This can be obtained by using the ideal gas equation as follow:
Volume (V) of O2 = 5.01 L
Pressure (P) = 0.860 atm
Number of mole (n) of O2 = 0.2685 mole
Gas constant (R) = 0.0821 atm.L/Kmol
Temperature (T) =..?
PV =nRT
0.860 x 5.01 = 0.2685 x 0.0821 x T
Divide both side by 0.2685 x 0.0821
T = (0.860 x 5.01) /(0.2685 x 0.0821)
T = 195.5 K
Therefore, the temperature of O2 must be 195.5 K.
At a certain temperature the equilibrium constant, Kc, equals 0.110 for the reaction: 2 ICl (g) ⇌ I2 (g) +Cl2 (g) What is the equilibrium concentration of ICl if 0.750 mol of I2 and 0.750 mol of Cl2 are initially mixed in a 1.00-L flask?
Answer:
The equilibrium concentration of ICl is 2.26 M
Explanation:
Chemical equilibrium is a state in which no changes are observed as time passes, despite the fact that the substances present continue to react. This is because chemical equilibrium is established when the forward and reverse reaction take place simultaneously at the same rate.
For the study of chemical equilibrium, the so-called equilibrium constant Kc is useful. Being:
aA + bB ⇔ cC + dD
the equilibrium constant Kc is:
[tex]Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }[/tex]
That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
For the reaction:
2 ICl (g) ⇌ I₂ (g) +Cl₂ (g)
the constant Kc is:
[tex]Kc=\frac{[I_{2} ]*[Cl_{2} ]}{[ICl]^{2} }[/tex]
Being Kc =0.110 and the concentration being the amount of moles of solute that appear dissolved in each liter of the mixture and being calculated by dividing the moles of the solute by the liters of the solution:
[tex][I_{2} ]=\frac{0.750 moles}{1 L} =0.750 \frac{moles}{L}[/tex][tex][Cl_{2} ]=\frac{0.750 moles}{1 L} =0.750 \frac{moles}{L}[/tex]and replacing in the constant we get:
[tex]0.110=\frac{0.750*0.750}{[ICl]^{2} }[/tex]
Solving, you get the ICl concentration at equilibrium:
[tex][ICl]^{2} =\frac{0.750*0.750}{0.110 }[/tex]
[tex][ICl] =\sqrt{\frac{0.750*0.750}{0.110 }}[/tex]
[ICl]= 2.26 M
The equilibrium concentration of ICl is 2.26 M
The solubility of calcium oxalate, CaC2O4, in pure water is 4.8 × 10‑5 moles per liter. Calculate the value of Ksp for silver carbonate from this data.
Answer:
2.3 × 10⁻⁹
Explanation:
Step 1: Write the reaction for the solution of calcium oxalate
CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)
Step 2: Make an ICE chart
We can relate the molar solubility (S) with the solubility product constant (Ksp) through an ICE chart.
CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)
I 0 0
C +S +S
E S S
The solubility product constant is:
Ksp = [Ca²⁺] × [C₂O₄²⁻] = S² = (4.8 × 10⁻⁵)² = 2.3 × 10⁻⁹
The substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an:_______
A) vicinal diol
B) geminal diol
C) acetal
D) ketal
Answer:
B) geminal diol
Explanation:
Hello,
In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.
Regards.
The mole is a counting number that allows scientists to describe how individual molecules and atoms react. If one mole of atoms or molecules is equal to 6.022 x 10^32 atoms or molecules, how many molecules are in 23.45 g sample of copper (II) hydroxide, Cu(OH)2? Express your answer to the correct number of significant figures. (MM of Cu(OH)2 is 97.562g/mol. Be sure to show all steps completed to arrive at the answer.
Answer:
[tex]\large \boxed{1.503 \times 10^{23}\text{ molecules of Cu(OH)}_{2}}$}[/tex]
Explanation:
You must calculate the moles of Cu(OH)₂, then convert to molecules of Cu(OH)₂.
1. Moles of Cu(OH)₂[tex]\text{Moles of Cu(OH)}_{2} = \text{24.35 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \text{0.2496 mol Cu(OH)}_{2}[/tex]
2. Molecules of Cu(OH)₂[tex]\text{No. of molecules} = \text{0.2496 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= \mathbf{1.503 \times 10^{23}}\textbf{ molecules Cu(OH)}_{2}\\\text{There are $\large \boxed{\mathbf{1.503 \times 10^{23}}\textbf{ molecules of Cu(OH)}_{2}}$}[/tex]
The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].
According to the Avogadro number, the number of molecules in a mole of atom has been equivalent to the Avogadro constant. The value of Avogadro constant has been [tex]\rm 6.023\;\times\;10^2^3[/tex].
The moles of a compound has been given as:
[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}[/tex]
The moles in 23.45 g copper (II) hydroxide has been:
[tex]\rm Moles=\dfrac{23.45}{97.562} \\Moles=0.24\;mol[/tex]
The moles of copper (II) hydroxide has been 0.24 mol.
The number of molecules in 0.24 mol sample has been driven by:
[tex]\rm 1\;mol=6.023\;\times\;10^2^3\;molecules\\0.24\;mol=0.24\;\times\;10^2^3\;molecules\\0.24\;mol=1.45\;\times\;10^2^3\;molecules[/tex]
The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].
For more information about molecules in a mole of sample, refer to the link:
https://brainly.com/question/24577356
The combination of a carbonyl group and a hydroxyl group on the same carbon atom is called a ________ group.
a. carbamate group
b. carbonate
c. carboxlate
d. carboxyl
Answer:
d. carboxyl
Explanation:
The presence of carbonyl group (>C=O)) and a hydroxyl group ( (−OH) on the same carbon atom is called a "carboxyl" group. A carboxyl group is represented as COOH and acts as the functional group part of carboxylic acids.
For example:
Formic acid or Methanoic acid (H-COOH) Butanoic acid (C3H7-COOH)Hence, the correct option is "d. carboxyl ".
Why are cells important to an organisms survival
Answer:
Cells are the basic structures of all living organisms. Cells provide structure for the body, take in nutrients from food and carry out important functions. ... These organelles carry out tasks such as making proteins?, processing chemicals and generating energy for the cell
Answer: I absolutely love this question! Biology is so interesting, so I always love to answer the curiosity of others regarding biology, such as that!
Cells are simply the basic structures of all organisms, that are living, of course! Cells provide structure for the body, and they also take in nutrients that your body needs from food and they carry out important functions. These organelles carry out tasks such as making proteins, processing chemicals, and generating energy for the cell. Isn’t that cool?
Hope this helped! <3
1.) A sample of neon gas at a pressure of 0.646 atm and a temperature of 242 °C, occupies a volume of 515 mL. If the gas is cooled at constant pressure until its volume is 407 mL, the temperature of the gas sample will be ________°C.
2.) A sample of argon gas at a pressure of 0.633 atm and a temperature of 261 °C, occupies a volume of 694 mL. If the gas is heated at constant pressure until its volume is 796 mL, the temperature of the gas sample will be___________°C.
3.) 0.962 mol sample of carbon dioxide gas at a temperature of 20.0 °C is found to occupy a volume of 21.5 liters. The pressure of this gas sample ismm ____________ Hg.
Answer:1 )T2=134°C 2) T2=339.48°C. 3)
P=817.59 mmHg.
Explanation:
1.Given ;
pressure, P1 of neon gas = 0.646 atm
temperature, T1 =242oC + 273=515oC
Volume, V1 =515ml
Volume V2= 407ml
temperature , T 2= ?
Solution;
And at constant pressure, the volume cools at V2=407 mL at T2=?
From ideal gas equation, PV=nRT
V/T=constant
therefore
V1/V2=T1/T2 = T2=(V2 xT1)/V1
T2=(407 mL x 515 K)/515 mL= 407K.
T2= 407K -273= 134°C. recall 0°C=273 K)
2..Given ;
pressure, P1 of neon gas = 0.633 atm
temperature, T1 =261oC + 273=534oC
Volume, V1 =694ml
Volume V2= 796ml
temperature , T 2= ?
Solution;
And at constant pressure, the volume expands at V2=796mL at T2=?
From ideal gas equation, PV=nRT
V/T=constant
therefore
V1/V2=T1/T2 = T2=(V2 xT1)/V1
T2=(796 mL x 534 K)/694mL= 612.48K.
T2= 612.48K -273= 339.48°C. recall 0°C=273 K
3
Given;
moles of CO2= n=0.962 mol,
temperature T=20°C=20+273 K =293 K,
volume V=21.5 L,
gas constant R at L·mmHg/mol·K= 62.3637 L mmHg mol^-1 K^-1
Using ideal gas equation PV=nRT
P=nRT/V
P=(0.962 mol)x(62.3637mmHg mol^-1 K^-1)x(293 K)/(21.5L)
P=817.59 mmHg.
Why can gasses change volume?
A. The forces holding the gas particles together are
stronger than gravity.
B. The gas particles have no mass, so they can change volume.
C. Gravity has no effect on gas particles, so they can float away.
O D. There are no forces holding the gas particles together.
Answer:
There are no forces holding the gas particles together.
Explanation:
A student completed the experiment but found that the total amount of material recovered weighed more than the original sample. What is the most likely source of error and how may it be corrected?
Answer:
This is due to the water moisture present in the recovered sample.
Explanation:
The total amount of material recovered isn’t meant to weigh more than the original sample. However when this happens then it means there is the presence of water moisture in the recovered sample.
The recovered samples however needs to be heated to make it dry and eliminate the water moisture through evaporation.
Amino acids can be synthesized by reductive amination. Draw the structure of the organic compound that you would use to synthesize serine.
Answer:
Following are the answer to this question:
Explanation:
For the reductive amination of its carbonyl group, amino acids could be synthesized by reducing ammunition, which can be synthesized in the given attachment file:
please find the attachment:
