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Setler79 [48]
3 years ago
7

If a lab requires each a lab group (3 students) to have 25 ml of a solution and it takes 15 grams of AgNO₃ cuprous nitrate, to m

ake 1 liter of solution, how many grams are needed to make enough solution?
Chemistry
1 answer:
pochemuha3 years ago
4 0

Answer:

0.375 grams are needed to make 25 mL solution.

Explanation:

Mass of AgNO_3 cuprous nitrate required to make 1 l of solution = 15 g.

1 L = 1000 mL

Mass of AgNO_3 cuprous nitrate required to make 1000 mL of solution = 15 g

Mass of AgNO_3 cuprous nitrate required to make 1 mL of solution:

=\frac{15}{1000} g

Mass of AgNO_3 cuprous nitrate required to make 25 mL of solution:

=\frac{15}{1000} \times 25 g=0.375 g

0.375 grams are needed to make 25 mL solution.

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Answer:

valence electrons are the electrons that are located in the outermost electron shell of an element. Knowing how to find the number of valence electrons in a particular atom is an important skill for chemists because this information determines the kinds of chemical bonds that it can form and, therefore, the element's reactivity

Explanation:

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In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 18.0°C. You identified that the ideal pressure (
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Answer:

4,38%

small molecular volumes

Decrease

Explanation:

The percent difference between the ideal and real gas is:

(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ <em>4,38%</em>

This difference is considered significant, and is best explained because argon atoms have relatively <em>small molecular volumes. </em>That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.

Therefore, an increasing in volume will produce an ideal gas behavior. Thus:

If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to <em>decrease</em>

<em />

I hope it helps!

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The percentage of keys you type correctly is A. accuracy B. correctness C. exactness D. precision WHO EVER GETS THIS EARNS BRAIN
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Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and hydrosulfuric aci
Helga [31]

Answer:

2OH-(aq) + 2H+(aq) → 2H2O(l)

Explanation:

Step 1: Data given

sodium hydroxide = NaOH

hydrosulfuric acid = H2S

Step 2: The unbalanced equation

NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

Step 3: Balancing the equation

On the left side we have 1x Na (in NaOH), on the right side we have 2x Na (in Na2S). To balance the amount of Na on both sides, we have to multiply NaOH on the left side by 2.

2NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

On the left side we have 4x H (2x in NaOH and 2x in H2S), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2. Now the equation is balanced.

2NaOH(aq) + H2S(aq) → Na2S(aq) + 2H2O(l)

Step 4: The net ionic equation

2Na+(aq) + 2OH-(aq) + 2H+(aq) + S^2-(aq) → 2Na+(aq) + S^2-(aq) + 2H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2OH-(aq) + 2H+(aq) → 2H2O(l)

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3 years ago
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