Dating of Rock Layers

A 3L sample of an ideal gas at 178 K has a pressure of 0.3 atm. Assuming that the volume is constant, what is the approximate pr

harina [27]

Answer: The approximate pressure of the gas after it is heated to 278 K is 0.468 atm.

Explanation:

Given: $T_{1}$ = 178 K, $P_{1}$ = 0.3 atm

$T_{2}$ = 278 K, $P_{2}$ = ?

According to Gay Lussac law, at constant volume the pressure of a gas is directly proportional to the temperature.

Formula used to calculate the pressure is as follows.

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\$

Substitute the values into above formula is as follows.

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\frac{0.3 atm}{178 K} = \frac{P_{2}}{278 K}\\P_{2} = 0.468 atm$

Thus, we can conclude that the approximate pressure of the gas after it is heated to 278 K is 0.468 atm.

3 0

2 years ago

A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.

Elodia [21]

Answer:

Hydration number: 4

Explanation:

1) Mass of water in the hydrated compound

Mass of water = Mass of the hydrated sample - mass of the dehydrated compound

Mass of water = 30.7 g - 22.9 g = 7.8 g

2) Number of moles of water

Number of moles = mass in grams / molar mass

molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol

Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol

3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂

The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g

Molar mass of Sr (NO₃)₂ : 211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).

Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol = 0.108 mol

4) Ratio

0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈ 4 mol H₂O : 1 mol Sr (NO₃)₂

Which means that the hydration number is 4.

4 0

3 years ago

Read 2 more answers

F Calculate the pH of 1.00 X 10^-6 M HCI *

amid [387]

Explanation:

A.ph 6 B. correct ph goes in a scale up to 14 below 7 is acidic above 7 is basic and in the middle is newtral when they say pH they are asking how acidic is it when they ask for pOH they are asking how basic it is pH is calculated using logs as is pOH so ans

12.65

5 0

3 years ago

The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and

Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

Moles KSCN = Moles Ag⁺ in the filtrate:

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

Total moles Ag⁺:

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

Moles of Ag⁺ in the precipitate:

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

Moles AsO₄ = Moles As:

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

Moles As₂O₃:

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

Mass As₂O₃:

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

7 0

3 years ago

URGENT CHEMISTRY EXPERT!

vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

$\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl$

(b) Moles of CuCl₂

$\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}$

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

$\text{Moles of Na$_{3}$PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}$

(d) Volume of Na₃PO₄

$V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}$

Part 2. Net ionic equation

(a) Molecular equation

$\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})$

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0

2 years ago