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zheka24 [161]
3 years ago
6

What volume (in L) of carbon dioxide will

Chemistry
1 answer:
Mademuasel [1]3 years ago
6 0
I need this answer as well. Please like this or comment when you find this out. Thank you :)
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How many molecules Of H20 are<br> equivalent to 97.2 g H20?<br> (H = 1.008 g/mol, O = 16.00 g/mol)
AleksandrR [38]

Answer:

m(H₂O) = 97,2 g.n(H₂O) = m(H₂O) ÷ M(H₂O).n(H₂O) = 97,2 g ÷ 18

Explanation:

8 0
2 years ago
A solution of H 2 SO 4 ( aq ) H2SO4(aq) with a molal concentration of 8.01 m 8.01 m has a density of 1.354 g / mL . 1.354 g/mL.
anzhelika [568]

Answer:

[H₂SO₄] = 6.07 M

Explanation:

Analyse the data given

8.01 m → 8.01 moles of solute in 1kg of solvent.

1.354 g/mL → Solution density

We convert the moles of solute to mass → 8.01 mol . 98g /1mol = 785.4 g

Mass of solvent = 1kg = 1000 g

Mass of solution = 1000g + 785.4 g = 1785.4 g

We apply density to determine the volume of solution

Density = Mass / volume → Volume = mass / density

1785.4 g / 1.354 g/mL = 1318.6 mL

We need this volume in L, in order to reach molarity:

1318.6 mL . 1L / 1000mL = 1.3186 L ≅ 1.32L

Molarity (mol/L) → 8.01 mol / 1.32L = 6.07M

4 0
3 years ago
Read 2 more answers
PLEASE HELP I WILL GIVE BRAINLYEST!!!!
adoni [48]
I think that it is true to the bone
8 0
3 years ago
Write and balance the equation for the combination of ammonia gas with solid copper (II) oxide to produce copper metal, nitrogen
Katena32 [7]

Explanation :

The balanced chemical reaction will be,

3CuO(s)+2NH_3(g)\rightarrow 3Cu(s)+N_2(g)+3H_2O(l)

By the stiochiometry, 3 moles of solid copper(II)oxide react with 2 moles of ammonia gas to give 3 moles of copper metal, 1 mole of nitrogen gas and 3 moles of liquid water.

The states of matter of each elements and compound is,

Copper(II)oxide is in solid state

Ammonia is in gaseous state

Copper metal is in solid state

Nitrogen is in gaseous state

Water is in liquid state

7 0
3 years ago
Read 2 more answers
Question 10
bixtya [17]
The reaction that should be followed is
Na2SO4 + C<span>a(NO3)2 --> CaSO4 + 2NaNO3</span>
first calculate the limiting reactant
mol Na2SO4 = 0.075 L (<span>1.54×10−2 mol / L) = 1.155x10-3 mol


mol Ca(NO3)2 = 0.075 L (</span><span>1.22×10−2 mol / L) = 9.15x10-4 mol
so the limiting reactant is the Ca(NO3)2

so all of the Ca2+ will be precipitated, percentage unprecipitated = 0.00 % </span>



5 0
4 years ago
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