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kolezko [41]
3 years ago
11

Gas

Chemistry
1 answer:
Schach [20]3 years ago
6 0

Answer:

this would be the .04 it is very simple

Explanation:

You might be interested in
How many moles of LiOH are needed to react completely with 25.5 g of CO2
LekaFEV [45]

Answer:

3.18 mol

Explanation:

2LiOH+CO_{2}-> Li_{2}CO_{3} +H_{2}O

n(CO2) = mass/ Mr.

             = 25.5 / 16

             = 1.59 mol

As per the equation above,

n(LiOH) : n(CO2)

     2      :    1

∴  3.18   :  1.59

     

3 0
2 years ago
What is the mass of 2.70 ×10^22 molecules of NaOH (Molar mass = 40.0 g/mol)?
Viefleur [7K]
Data:
Molar Mass of NaOH = 40 g/mol

Solving: <span>According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules
</span>
1 mol -------------------- 6.02*10²³ molecules
y mol -------------------- 2.70*10²² molecules

6.02*10²³y = 0.270*10²³ 
y =  \frac{0.270*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}{6.02*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}
\boxed{y \approx 0.04\:mol}


Solving: <span>Find the mass value now
</span>
40 g ----------------- 1 mol of NaOH
x g ------------- 0.04 mol of NaOH

x = 40*0.04
\boxed{\boxed{x = 1.6\:g}}\end{array}}\qquad\quad\checkmark

Answer:
The mass is 1.6 grams
6 0
3 years ago
How many L are in 1,500cm^3<br><img src="https://tex.z-dn.net/?f=1500%7Bcm%7D%5E%7B3%7D%20%3D%20%5C%3A%20...%20%20%5C%3A%20liter
irga5000 [103]
1.5 liters are in 1,500cm^3.
8 0
3 years ago
M=<br> V=432.6 mL<br> D=8.4 g/ml
Nitella [24]

Answer:

Explanation:

M=D times V

Answer-3,633.84g

Rounded Answer (correct sig figs)- 3600g

6 0
3 years ago
A total of 25.0 mL of 0.150 M potassium hydroxide (KOH) was required to neutralize 15.0 mL of sulfuric acid (H2SO4) of unknown c
Kamila [148]
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:

KOH + H₂SO₄ → H₂O + KHSO₄

If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:

0.025 L x 0.150 mol/L = .00375 mol KOH

0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄

We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:

0.00375 mol / 0.015 L = 0.25 mol/L

The concentration of H₂SO₄ being neutralized is 0.25 M.
6 0
3 years ago
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