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777dan777 [17]
3 years ago
10

2.0 kilograms of lithium hydroxide can “scrub”, or remove, how many liters of carbon dioxide at standard conditions?

Chemistry
1 answer:
Murljashka [212]3 years ago
7 0

Answer:

0.94 L

Explanation:

The equation of the reaction is;

2LiOH + CO2 -----> Li2CO3 + H2O

Number of moles LiOH = 2.0/23.95 g/mol = 0.084 moles

2 moles of LiOH scrubs 22.4 L of CO2

0.084 moles of LiOH scrubs 0.084 moles × 22.4/2

= 0.94 L

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Ethylene glycol (C2H6O2) is used as an antifreeze in cars. If 250 g of ethylene glycol is added to 3.00 kg of water, what is the
Zepler [3.9K]

Answer:

2,909 M

Explanation:

molair mass is of.ethylene is 26,04 g/mol

first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.

3000 g x 1.1 = 3300 mL = 3,3 L

Next you need to calculate the amount of moles:

250 g / 26,04 g/mol = 9,60 mol

Now you can calculate the molarity:

9,6/3.3 = 2,909 M

I don't know the answer for the second question. I'm sorry.

3 0
3 years ago
What mass of propane could burn in 48.0 g of oxygen? C3H8 + 5O2 → 3CO2 + 4H2O
Ipatiy [6.2K]

Answer:

Mass = 13.23 g  

Explanation:

Given data:

Mass of oxygen = 48.0 g

Mass of propane burn = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂     →      3CO₂ + 4H₂O

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 48.0 g/ 32 g/mol

Number of moles = 1.5 mol

now we will compare the moles of propane and oxygen.

              O₂           :          C₃H₈

               5            :            1

             1.5            :          1/5×1.5 = 0.3 mol

Mass of propane burn:

Mass = number of moles × molar mass

Mass = 0.3 mol × 44.1 g/mol

Mass = 13.23 g  

6 0
3 years ago
Pls help
bezimeni [28]

Answer:

Particle Symbol Mass

electron e- 0.0005486 amu

proton p+ 1.007276 amu

neutron no 1.008665

8 0
3 years ago
2. The density of helium is 1.78 X 104 g/cm. What is this<br> density in Dg/um??
Zepler [3.9K]

Answer:

d=1.78\times 10^{-7}\ Dg/\mu m^3

Explanation:

Given,

The density of Helium is 1.78\times 10^4\ g/cm^3

We need to find the density in Dg/μm

We know that,

1 g = 10 dg

1 cm³ = 10¹² μm³

So,

d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3

So, the density of Helium is equal to 1.78\times 10^{-7}\ Dg/\mu m^3.

4 0
2 years ago
Plssssssssssssssssssmejepleplele
Leona [35]

Answer:

Yes the two of the answer is True

3 0
3 years ago
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