Can someone please help me pls

If 5.0 liters H2 (g) at STP is heated to a temperature of 985, pressure remaining constant, the new volume of the gas will be?

vlada-n [284]

Answer:

$V_2=18 \ L \ H_2$

General Formulas and Concepts:

Chemistry - Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Charles' Law: $\frac{V_1}{T_1} =\frac{V_2}{T_2}$

Explanation:

Step 1: Define

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

Step 2: Solve for new volume

Substitute: $\frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}$
Cross-multiply: $(5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)$
Multiply: $4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K$
Isolate x: $18.0403 \ L \ H_2 = x$
Rewrite: $x=18.0403 \ L \ H_2$

Step 3: Check

We are given 2 sig figs as the smallest. Follow sig fig rules and round.

$18.0403 \ L \ H_2 \approx 18 \ L \ H_2$

5 0

3 years ago

Help me find the word elliptical

maria [59]

Omg where is it?? i’ve been looking for a long time

7 0

3 years ago

What is electron shielding

Artist 52 [7]

Shielding electrons are the electrons in the energy levels between the nucleus and the valence electrons. They are called "shielding" electrons because they "shield" the valence electrons from the force of attraction exerted by the positive charge in the nucleus. Hope this helps!!

8 0

3 years ago

Read 2 more answers

Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find

uranmaximum [27]

Answer:

The molecular formula = $C_2Cl_{4}F_2$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 11.79

Molar mass of C = 12.0107 g/mol

% moles of C = 11.79 / 12.0107 = 0.9816

% of Cl = 69.57

Molar mass of Cl = 35.453 g/mol

% moles of Cl = 69.57 / 35.453 = 1.9623

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64

Molar mass of F = 18.998 g/mol

% moles of F = 18.64 / 18.998 = 0.9812

Taking the simplest ratio for C, Cl and F as:

0.9816 : 1.9623 : 0.9812

= 1 : 2 : 1

The empirical formula is = $CCl_2F$

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

Volume = 458 mL = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 0.00052445 moles

Given that :

Amount = 0.107 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.00052445= \frac{0.107\ g}{Molar\ mass}$

$Molar\ mass= 204.0233\ g/mol$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

The molecular formula = $C_2Cl_{4}F_2$

6 0

4 years ago

What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 123 g) is decomposed to produce 3.2 g of O2 (molar mass 32 g)

My name is Ann [436]

Answer:

The percentage yield of O2 is 66.7%

Explanation:

Reaction for decomposition of potassium chlorate is:

2KClO₃ → 2KCl + 3O₂

The products are potassium chloride and oxygen.

Let's find out the moles of chlorate.

Mass / Molar mass = Moles

12.3 g / 123 g/mol = 0.1 mol

So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.

Then, 0.1 mol of chlorate may produce (0.1 .3)/ 2 = 0.15 moles

Let's convert the moles of produced oxygen, as to find out the theoretical yield.

0.15 mol . 32 g/ 1mol = 4.8 g

To calculate the percentage yield, the formula is

(Produced Yield / Theoretical yield) . 100 =

(3.2g / 4.8g) . 100 = 66.7 %

8 0

4 years ago