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algol13
2 years ago
14

Mixing agno3 and nacl will produce what white precipitate.

Chemistry
1 answer:
SSSSS [86.1K]2 years ago
3 0
AgCl silver chloride
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If a sample of air initially occupies 240L at 2 atm how much pressure is required to compress it to 20L at constant temperature
IceJOKER [234]

Answer:

24 atm.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 240 L

Initial pressure (P₁) = 2 atm

Final volume (V₂) = 20 L

Temperature = constant

Final pressure (P₂) =?

The final pressure required, can be obtained by using the Boyle's law equation as shown below:

P₁V₁ = P₂V₂

2 × 240 = P₂ × 20

480 = P₂ × 20

Divide both side by 20

P₂ = 480 / 20

P₂ = 24 atm

Thus, the final pressure required is 24 atm.

4 0
3 years ago
Why are the outer-most electrons the only ones included in the electron dot<br> diagram?
kupik [55]

Answer:

Atoms with 5 or more valence electrons gain electrons forming a negative ion, or anion. why are outermost electrons only ones included in orbital filling diagram? they are the only ones involved in chemical reactions and bonding. ... 2s orbital is farther from the nucleus meaning it has more energy.

Explanation:

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3 years ago
How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac
Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

,

• using the molar mass of Ca(OH)2 (74g/mol).

,

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

,

• using the molar mass of HNO3 (63.02g/mol).

4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3

So, 7.32g of HNO3 are required.

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1 year ago
First to answer both CORRECTLY gets brainleist!
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