Thermal decomposition of magnesium nitrate and magnesium hydroxide

How many Liters of space will a 70.0g sample of CO2 occupy?

Tanzania [10]

Answer:

35.6 liters at STP

Explanation:

The molar mass of carbon dioxide is about 44.01 g/mol. The volume of a mole of ideal gas at STP is 22.4 L, so the volume of 70.0 g will be ...

(70.0g)/(44.01 g/mol)·(22.4 L/mol) ≈ 35.6 L

5 0

3 years ago

How many CaF are in a 1.7x10^25 please I need help fast!!

Gemiola [76]

<h3>Answer:</h3>

28 mol CaF

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[Given] 1.7 × 10²⁵ molecules CaF

[Solve] moles CaF

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

[DA] Set up: $\displaystyle 1.7 \cdot 10^{25} \ molecules \ CaF(\frac{1 \ mol \ CaF}{6.022 \cdot 10^{23} \ molecules \ CaF})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 28.2298 \ moles \ CaF$

Step 4: Check

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

28.2298 mol CaF ≈ 28 mol CaF

7 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 ml at a temperature 126 °c and a pressure of 777

AleksAgata [21]

Answer : 72.05 g/mol

Explanation : 

Let's assume that the given gas is an ideal gas. Then we can use ideal gas equation,
PV = nRT

Where,
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol⁻¹ K⁻¹)
T = temperature in Kelvin (K)

The given data for the gas is,
P = 777 torr = 103591 Pa
V = 125 mL = 125 x 10⁻⁶ m³
T = (126 + 273) = 399 K
R = 8.314 J mol⁻¹ K⁻¹
n = ?

By applying the formula,
103591 Pa x 125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K
n = 3.90 x 10⁻³ mol

Moles (mol) = mass (g) / molar mass (g/mol)

Mass of the gas = 0.281 g
Moles of the gas = 3.90 x 10⁻³ mol
Hence,
molar mass of the gas = mass / moles
= 0.281 g / 3.90 x 10⁻³ mol
 = 72.05 g/mol

7 0

3 years ago

Read 2 more answers

When adding or subtracting deelmals, how many digits should the answer contain?

Alexus [3.1K]

Answer:

Depends, but in most cases, 2.

It's best to use as many digits as possible to keep it accurate.

Explanation:

This varies between teachers, as most schools go with 2 decimal places.

This is something that depends in your situation.

You technically want as many decimals as possible to keep it as accurate, but most people stick with 2.

I personally do 3, and commonly do 5 sometimes.

5 0

3 years ago

Thermal decomposition of magnesium nitrate and magnesium hydroxide ​

Thermal decomposition of magnesium nitrate and magnesium hydroxide