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Alja [10]
3 years ago
13

Please answer the question

Chemistry
2 answers:
slamgirl [31]3 years ago
8 0

this is a very hard question

qwelly [4]3 years ago
8 0
We know that he was in hamilton
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What is the concentration of NaCI in an aqueous solution that contains 0.032 grams of NaCI in 600. Grams of the solution
jolli1 [7]

Answer:

0.00032 Grams of NaCl per 1 gram of the solution

Explanation:

8 0
3 years ago
1.2 L sample of gas is determined to contain 0.5 moles of nitrogen. At the same temperature and pressure, how many moles of gas
hram777 [196]

A 20 L sample of the gas contains 8.3 mol N₂.

According to <em>Avogadro’s Law,</em> if <em>p</em> and <em>T</em> are constant

<em>V</em>₂/<em>V</em>₁ = <em>n</em>₂/<em>n</em>₁

<em>n</em>₂ = <em>n</em>₁ × <em>V</em>₂/<em>V</em>₁

___________

<em>n</em>₁ = 0.5 mol; <em>V</em>₁ = 1.2 L

<em>n</em>₂ = ?;           <em>V</em>₂ = 20 L

∴<em>n</em>₂ = 0.5 mol × (20 L/1.2 L) = 8.3 mol

4 0
3 years ago
en un recipiente se tiene 800 g de una solución al 35% en masa de ácido sulfuroso, de la cual se evapora 80ml de agua. ¿cuál es
Alinara [238K]

The mass percent of sulfurous acid in the new solution : 38.9%

<h3>Further explanation</h3>

<em>In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.</em>

<em />

solution 1

composition :

  • 35% acid :

\tt 0.35\times 800~g=280~g

  • water :

\tt 800-280=520~g

solution 2(new solution)

composition :

  • water

\tt 520-(80~ml\times 1~g/ml)=440~g

  • Total mass of new solution after water evaporated

\tt 280(acid)+440(water)=720~g

  • %mass of acid in a new solution

\tt \dfrac{280}{720}\times 100\%=38.9\%

5 0
2 years ago
Use the reaction given below to solve the problem that follows: Calculate the mass in grams of aluminum oxide produced by the re
bearhunter [10]

Answer:  28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}   

\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles

The balanced chemical equuation is:

4Al+3O_2\rightarrow 2Al_2O_3  

According to stoichiometry :

4 moles of Al produce == 2 moles of Al_2O_3

Thus 0.556 moles of Al will produce=\frac{2}{4}\times 0.556=0.278moles  of Al_2O_3

Mass of Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g

Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.

7 0
2 years ago
Fill in the Blank
Arisa [49]

Answer: Quantitative data

Explanation:

3 0
3 years ago
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