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mariarad [96]
3 years ago
7

1) Alarge doud of dust and gas spread out in a large area is

Chemistry
1 answer:
alexdok [17]3 years ago
8 0

Answer:

D. nebula

Explanation:

Hope this helps

<3

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Consider the following reaction. Zn(s) 2agno3(aq) --&gt; 2ag(s) zn(no3)2(aq)when 16. 2 g of silver was produced, _____ mole(s) o
Crazy boy [7]

The required amount of silver nitrate to produce 16.2g of silver is 25.48 grams.

<h3>What is the relation between mass & moles?</h3>

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

  • W = given mass
  • M = molar mass

Moles of silver = 16.2g / 107.8g/mol = 0.15mol

From the stoichiometry of the given reaction it is clear that, same moles of silver nitrate is required to produce same moles of silver. So 0.15 moles of silver nitrate is required.

Mass of silver nitrate = (0.15mol)(169.87g/mol) = 25.48g

Hence required mass of silver nitrate is 25.48g.

To know more about mass & moles, visit the below link:

brainly.com/question/19784089

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5 0
2 years ago
What is the mass measurement shown on the scales of this balance?
otez555 [7]
The answer is definitely not 70.923
6 0
3 years ago
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A student places a beaker on a hot plate. To the beaker, he adds a clear liquid and a white powder. The mixture instantly begins
Elenna [48]
True, because if it wasn't a chemical reaction it would have proceeded to stay the same. but it begins to bubble.
sorry if this isn't the best answer I'm trying my best.
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4 years ago
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How old was the oldest animal fossil <br>help thx​
JulijaS [17]

Answer:

the Rhyniognatha hirsti

Explanation:

at age 400 million years old

5 0
3 years ago
Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in Part A, and the temperature and
andrew11 [14]

Explanation:

Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.

This means that, for N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}

 moles of N_{2} + moles of H_{2} = 0.908 mol

Since, 2 moles of N_{2} = 2 \times 0.227 = 0.454 mol

As it is known that the ideal gas equation  is PV = nRT

And, as the temperature and volume were kept constant, so we can write

        \frac{P(_in)}{n_(in)} = \frac{P_(final)}{n_(final)}

          \frac{10.4}{0.908} = \frac{P_(final)}{0.454&#10;}

       P_(final) = 10.4 \times \frac{0.454}{0.908}

                            = 5.2 atm

Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.

7 0
3 years ago
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