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3 years ago

How many grams are in 0.35 moles of C2H4? please show dimensional analysis

Chemistry

1 answer:

fredd [130]3 years ago

5 0

Answer:

9.8g

Explanation:

Using periodic table find molar mass of C and H:

C=12.01g/mol

H=1.008g/mol

Molar mass of C2H4=(12.01)2+(1.008)4=28.03g/mol

Using molar mass times moles of the chemical to find the mass in 0.35 moles of c2h4:

28.03 x 0.35=9.8grams

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3 years ago

Calculate the number of kilojoules to warm 125 g of iron from 23.5 °C to 78.0 °C.

insens350 [35]

Given:
Iron, 125 grams T
1 = 23.5 degrees Celsius, T2 = 78 degrees Celsius.

Required:
Heat produced in kilojoules

Solution:
The molar mass of iron is 55.8 grams per mole. SO we need to change the given mass of iron into moles.

Number of moles of iron = 125 g/(55.8 g/mol) = 2.24 moles

Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2.24 moles (8.314 Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
Q (heat) = 1014.97 Joules or 1.015 kilojoules This is the amount of heat produced in warming 125 g f iron.

7 0

3 years ago

the gas left in an used aerosol can is at a pressure of 103 kPa at 25 degrees celsius if this can be thrown into fire what is th

Rainbow [258]

Hello!

The pressure of the gas when it's temperature reaches 928 °C is 3823,36 kPa

To solve that we need to apply Gay-Lussac's Law. It states that the pressure of a gas when the volume is left constant (like in the case of a sealed container like an aerosol can) is proportional to temperature. This is the relationship derived from this law that we use to solve this problem:

$P2= \frac{P1}{T1}*T2= \frac{103 kPa}{25}*928=3823,36 kPa$

Have a nice day!

5 0

3 years ago

Read 2 more answers

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr

umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0

3 years ago

What actions show conserving by reusing? Choose the two correct answers

hram777 [196]

A and D

Hope this helps

3 0

3 years ago