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Maru [420]
3 years ago
6

My insta is gageisgod7j7 HMU

Chemistry
2 answers:
Tcecarenko [31]3 years ago
7 0
Alright sounds good and also thank you so much for the free points dawg
PilotLPTM [1.2K]3 years ago
5 0

Answer:

Maybe ;) maybe not hmmm

Explanation:

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A gas has density 2.41 g/liter at 25°C and 770 mm Hg. Calculate it's molecular mass (R = 0.0821 L atm.mol-1K-1 ​
Evgesh-ka [11]

Answer:

Explanation:

Given : Density - 2.41 g/liter  

Temperature - 25° C  

Pressure : 770 mm Hg  

R = 0.0821 L atm mol-¹K-¹

 

Find : Molecular mass of gas

Solution : Ideal gas equation with respect to density will be : PM = dRT. In the formula, P is pressure, M is molecular mass, d is density, R is gas constant and T is temperature.

Keeping the values in equation-  

Pressure : 770 mm Hg = 1 atm  

Temperature : 273 + 25 = 298 K

 

M = dRT/P  

M = (2.41*0.0821*298)/1  

M = 58.96 gram/mol

Thus, the molecular mass of gas is 58.96 gram/mol.

8 0
3 years ago
Matthew was working with different concentrations of hydrochloric acid in the lab. Which of these would BEST describe the result
nevsk [136]

The answer to this question would be C. I hope this helps!

7 0
3 years ago
Read 2 more answers
What is the volume occupied by 30.00 grams of pure gold?​
mr Goodwill [35]

Answer:

I think 1.56

Explanation:

3 0
3 years ago
How many liters of hydrogen are required to react completely with 2.4L of oxygen to form water? 2H2 + O2 --> 2H2O
RoseWind [281]

Answer:

2.4 mole of oxygen will react with 2.4 moles of hydrogen

Explanation:

As we know

1 liter = 1000 grams

2H2 + O2 --> 2H2O

Weight of H2 molecule = 2.016 g/mol

Weight of water = 18.01 gram /l

2 mole of oxygen react with 2 mole of H2

2.4 mole of oxygen will react with 2.4 moles of hydrogen

3 0
2 years ago
What is the mass of a sample of water that takes 2000 kJ of energy to boil into steam at 373 K. The latent heat of vaporization
zzz [600]

Answer:

\boxed{\text{889 g}}

Explanation:

The formula relating the mass m of a sample and the heat q to vaporize it is

q = mL, where L is the latent heat of vaporization.

\begin{array}{rcl}2000 \times 10^{3} \text{ J} & = & m \times \dfrac{2.25 \times 10^{6} \text{ J}}{\text{1 kg}}\\\\m & = & \dfrac{2000 \times \times 10^{3}\text{ kg}}{2.25 \times 10^{6}}\\ & = & \text{0.889 kg}\\\\ & = & \text{889 g}\\\end{array}\\\text{The mass of water is $\boxed{\textbf{889 g}}$}

5 0
3 years ago
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