Is neon a metal, nonmetal , or metalloid

What is the mass in grams of 5.00moles of CH4?

anastassius [24]

Answer:

1. 80g

2. 1.188mole

Explanation:

1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Number of mole of CH4 from the question = 5 moles

Mass of CH4 =?

Mass = number of mole x molar Mass

Mass of CH4 = 5 x 16

Mass of CH4 = 80g

2. Mass of O2 from the question = 38g

Molar Mass of O2 = 16x2 = 32g/mol

Number of mole O2 =?

Number of mole = Mass /Molar Mass

Number of mole of O2 = 38/32

Number of mole of O2 = 1.188mole

6 0

3 years ago

Both the physical and chemical properites of a substance can change when a substance undergoes a chemical reaction. When a log i

ElenaW [278]

Answer is option C...

6 0

3 years ago

Read 2 more answers

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

An engine operates on a Carnot cycle that uses 1mole of an ideal gas as the

sattari [20]

Answer:

Step 1;

q = w = -0.52571 kJ, ΔS = 0.876 J/K

Step 2

q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ

Explanation:

The given parameters are;

$P_i$ = 100 N·m

$T_i$ = 327 K

$P_f$ = 90 N·m

Step 1

For isothermal expansion, we have;

ΔU = ΔH = 0

w = n·R·T·ln( $P_f$ / $P_i$ ) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71

w ≈ -0.52571 kJ

At state 1, q = w = -0.52571 kJ

ΔS = -n·R·ln( $P_f$ / $P_i$ ) = -1 × 8.314 × ln(90/100) ≈ 0.876

ΔS ≈ 0.876 J/K

Step 2

q = 0 for adiabatic process

ΔU = 25×(27 - 327) = -7,500

w = ΔU = -7.5 kJ

ΔH = ΔU + n·R·ΔT

ΔH = -7,500 + 8.3142 × 300 = -5,005.74

ΔH = ΔU = -5.00574 kJ

6 0

3 years ago

Identify this symbol. Cl-

g100num [7]

Symbols on the periodic table represent an element.

Cl is also known as chlorine.

8 0

3 years ago