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3 years ago

What is the volume in liters of 3.39 kg of ethylene glycol?

Chemistry

1 answer:

Sergeu [11.5K]3 years ago

4 0

The volume in liters is v=3.7xL

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If an ideal gas has a pressure of 5.49 atm, a temperature of 88.78 °c, and has a volume of 22.03 l, how many moles of gas are in

kirza4 [7]

N = (PV)/RT
(T = 88.78 + 273 = 361.78K)
(R = 22.4/273 = 0.082)
= (5.49 x 22.03)/(0.082 x 361.78) = ?
Put it into the calculator. It's hard to do that on a mobile phone.

7 0

4 years ago

CH4 + 202 → CO2 + 2H2O

ohaa [14]

It’s a combustion reaction but what are you trying to find?

6 0

2 years ago

3. What was the temperature in

Alexeev081 [22]

Q = mcΔT

m = 75 g
c = 4.25 J/gram°C
q = 2450 J

q/mc = ΔT

2450/(75 • 4.25) = 7.7°C

The temperature increased by 7.7°C.

6 0

2 years ago

Oxalic acid can remove rust (Fe2O3) caused by bathtub rings according to the reaction Fe2O3(s) - 6H2C2O4(aq) rightarrow 2Fe(C2O4

Katena32 [7]

Answer: The mass of rust that can be removed is 1.597 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of oxalic acid solution = 0.1255 M

Volume of solution = $6.00\times 10^2mL$ = 600 mL = 0.600 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of oxalic acid}}{0.600L}\\\\\text{Moles of oxalic acid}=(0.100mol/L\times 0.600L)=0.06mol$

For the given chemical reaction:

$Fe_2O_3(s)+6H_2C_2O_4(aq.)\rightarrow 2Fe(C_2O_4)_3^{3-}(aq.)+3H_2O(l)+6H^+(aq.)$

By Stoichiometry of the reaction:

6 moles of oxalic acid reacts with 1 mole of ferric oxide (rust)

So, 0.06 moles of oxalic acid will react with = $\frac{1}{6}\times 0.06=0.01mol$ of ferric oxide (rust)

To calculate the mass of rust for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of rust (ferric oxide) = 159.7 g/mol

Moles of rust = 0.01 moles

Putting values in above equation, we get:

$0.01mol=\frac{\text{Mass of rust}}{159.7g/mol}\\\\\text{Mass of rust}=(0.01mol\times 159.7g/mol)=1.597g$

Hence, the mass of rust that can be removed is 1.597 grams

5 0

3 years ago

How many grams of copper are needed to produce 53 grams of silver in this single

Mars2501 [29]

Answer: 16g Cu or 15.6g Cu

Explanation:

7 0

3 years ago