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Lelechka [254]
2 years ago
5

Can u help me with this

Chemistry
1 answer:
dolphi86 [110]2 years ago
8 0

Answer:

c

Explanation:

the reason is because your a homosiejenenddj

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Someone help answer this for me tysm
Ghella [55]
Energy, kinetic, potential
8 0
3 years ago
Read 2 more answers
A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga
Elden [556K]

Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

molecular wt of unknown gas=8.102/0.0806=100.52

8 0
3 years ago
3. In the formation of acid rain, sulfur dioxide reacts with oxygen and water in the air to form sulfuric acid. Write a balanced
Andreyy89

Answer:

SO₂ + 0.5 O₂ + H₂O → H₂SO₄

3.83 g

Explanation:

In the formation of acid rain, sulfur dioxide reacts with oxygen and water in the air to form sulfuric acid. The balanced chemical equation is:

SO₂ + 0.5 O₂ + H₂O → H₂SO₄

The molar mass of SO₂ is 64.07 g/mol. The moles of SO₂ corresponding to 2.50 g are:

2.50 g × (1 mol/64.07 g) = 0.0390 mol

The molar ratio of SO₂ to H₂SO₄ is 1:1. The moles of H₂SO₄ formed are 0.0390 moles.

The molar mass of H₂SO₄ is 98.08 g/mol. The mass of H₂SO₄ is:

0.0390 mol × 98.08 g/mol = 3.83 g

4 0
3 years ago
Where is earths mantle located
matrenka [14]
<span>it is located directly under the sima</span>
4 0
3 years ago
Read 2 more answers
How many grams of PH3 will be produced if 6.0 l of H2 are used?
elena-s [515]

Mass of PH3= 6.086 g

<h3>Further explanation</h3>

Given

6.0 L of H2

Required

mass of PH3

Solution

Reaction

P4 + 6H2 → 4PH3

Assumed at STP ( 1 mol gas=22.4 L)

Mol of H2 for 6 L :

= 6 : 22.4 L

= 0.268

From the equation, mol PH3 :

= 4/6 x moles H2

= 4/6 x 0.268

= 0.179

Mass PH3 :

= 0.179 x 33,99758 g/mol

= 6.086 g

5 0
2 years ago
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