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Ilya [14]
2 years ago
4

How well can you apply Boyle’s law to this sample of gas that experiences changes in pressure and volume? Assume that temperatur

e and number of moles of gas are constant in this problem.
Using the first volume and pressure reading on the table as V1 and P1, solve for the unknown values in the table below. Remember to use the rules of significant figures when entering your numeric response.
Chemistry
1 answer:
BabaBlast [244]2 years ago
4 0

Answer:

The answers are:

A = 1 L

B = 0.5 atm

C = 0.6 atm

D = 4 L

Explanation:

Boyle's Law states that if the temperature and number of moles of gas are constant, then pressure (P) and volume (V) vary according to the relation:

PV = constant

To find A:

P_{1}V_{1}=P_{2}V_{2}\\1.5\times2.0=3.0\times A\\A=1L\\

To find B:

P_{1}V_{1}=P_{2}V_{2}\\1.5\times2=B\times6\\B=0.5\:atm\\

To find C:

P_{1}V_{1}=P_{2}V_{2}\\1.5\times2.0=C\times5.0\\C=0.6\:atm

To find D:

P_{1}V_{1}=P_{2}V_{2}\\1.5\times2.0=0.75\times D\\D=4L

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Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As
Crank

Answer:

1.8 × 10⁻¹⁶ mol  

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

                         1.2 + 3s          2s

K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} =  1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0
2 years ago
mixture or pure substance: 1.blood 2.dyes 3.self-raising flour 4.muesli 5.copper wire 6.distilled water 7.table salt 8.milk 9.br
jenyasd209 [6]

Answer: Mixture: Blood , Self raising flour,muesli ,dyes, milk, tea, air, bronze

Pure substance: Copper wire, distilled water, table salt, oxygen.

Explanation:

Mixture is a substance which is made up two or more number of compounds which chemically inactive and retain their distinct chemical properties.

Blood , Self raising flour,muesli ,dyes, milk, tea, air, bronze

Pure substance is defined as anything with uniform and unchanging composition is known s pure substance.

Copper wire, distilled water, table salt, oxygen.

5 0
2 years ago
A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete form
liberstina [14]

Explanation:

Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.

Mass of sample (CuSO4 * nH2O)

= 152.00g - 128.10g = 23.90g.

Mass of water loss during heating

= 152.00g - 147.60g = 4.40g.

Molar mass of H2O = 18g/mol

Moles of H2O in sample

= 4.40g / (18g/mol) = 0.244mol.

Mass of anhydrous sample (CuSO4)

= 23.90g - 4.40g = 19.50g

Molar mass of CuSO4 = 159.61g/mol

Moles of CuSO4 in sample

= 19.50g / (159.61g/mol) = 0.122mol.

Since mole ratio of CuSO4 to H2O

= 0.122mol : 0.244mol = 1:2, n = 2.

Hence we have CuSO4 * 2H2O.

6 0
2 years ago
Read 2 more answers
Determine the poh of a 0.348 m ba(oh)2 solution at 25°c.
vladimir1956 [14]

<u>Given:</u>

Concentration of Ba(OH)2 = 0.348 M

<u>To determine:</u>

pOH of the above solution

<u>Explanation:</u>

Based on the stoichiometry-

1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion

Therefore, concentration of OH- ion = 2*0.348 = 0.696 M

pOH = -log[OH-] = - log[0.696] = 0.157

Ans: pOH of 0.348M Ba(OH)2 is 0.157

6 0
2 years ago
Read 2 more answers
When dissolved beryllium chloride reacts with dissolved silver nitrate in water, aqueous beryllium nitrate and silver
lisabon 2012 [21]

Answer:

The balanced chemical equation is :

BeCl_{2} + 2AgNO_{3}\rightarrow Be(NO_{3})_{2} + 2AgCl

Explanation:

The chemical equation in which number of atoms in reactants is equal to products is called balanced equation.

Formula of , Beryllium Chloride :[BeCl_{2}

Beryllium Nitrate : BeNO_{3}

Silver Chloride : AgCl

Silver Nitrate  :AgNO_{3}

When beryllium chloride reacts with dissolved silver nitrate in water , following reaction occur :

 BeCl_{2} + 2AgNO_{3}\rightarrow Be(NO_{3})_{2} + 2AgCl

The number of atoms in reactant as well as in products are balanced :

Be = 1

Ag = 2

N =2

O = 6

Cl = 2

4 0
2 years ago
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