How many moles of KNO3 are in 500 mL of 2.0 M KNO3? __

What could you do to change the volume of a gas

Jet001 [13]

Solids and liquids have volumes thatdo not change easily. A gas, on the other hand, has a vol- ume thatchanges to match the volume of its container. The molecules in a gas are very far apart compared with the molecules in a solid or a liquid. The amount of space between the molecules in a gas can change easily.

5 0

3 years ago

Which balanced equation represents a redox reaction? a. AgNO3(aq) +NaCl(aq) →AgCl(s) +NaNO3(aq) b. H2CO3(aq) →H2O() + CO2(g) c.

Aleks [24]

Answer : The correct option is, (D) $Mg(s)+2HCl(aq)\rightarrow MgCl_2(s)+H_2(g)$

Explanation :

(A) $AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

This reaction is a double displacement reaction in which the cation and anion of two reactants are exchange their places to give two different products.

(B) $H_2CO_3(aq)\rightarrow H_2O(l)+CO_2(g)$

This reaction is a decomposition reaction in which the larger molecule decomposes to give two or more products.

(C) $NaOH(aq)+HCl(aq)\rightarrow NaCl(s)+H_2O(aq)$

This reaction is a neutralization reaction in which an acid and a base react to give a salt and water as a product.

(D) $Mg(s)+2HCl(aq)\rightarrow MgCl_2(s)+H_2(g)$

This reaction is a redox reaction in which the oxidation and reduction reaction occur simultaneously.

Oxidation reaction is the reaction in which a substance looses its electrons. In this oxidation state increases.

Reduction reaction is the reaction in which a substance gains electrons. In this oxidation state decreases.

In this reaction, magnesium shows oxidation due to change in oxidation number from (0) to (+2) and hydrogen shows reduction due to change in oxidation number from (-1) to (0).

Hence, the correct option is, (D)

7 0

3 years ago

When heated, mercury oxide produces oxygen plus mercury. What would be the combined mass of oxygen and mercury if 20g of mercury

vivado [14]

Releasing the oxygen from the mercury does not change the mass. If you started with 20 g total you would still have 20 g total after the heating

7 0

3 years ago

Which of the following aqueous solutions are good buffer systems?

garri49 [273]

The following aqueous solutions represents good buffer systems:

0.22 M acetic acid + 0.15 M potassium acetate
0.29 M ammonium nitrate + 0.32 M ammonia

<h3>What is a buffer?</h3>

A buffer is a solution used to stabilize the pH (acidity) of a liquid.

A good buffer system is generally known to contain close or equal concentrations of a weak acid and its conjugate base.

Based on the above explanation, the following represents a good buffer system as they are between their weak acid and conjugate base:

0.22 M acetic acid + 0.15 M potassium acetate
0.29 M ammonium nitrate + 0.32 M ammonia

Learn more about buffer at: brainly.com/question/22821585

#SPJ1

8 0

1 year ago

Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine

Yuliya22 [10]

Answer:

194 g/mol

Explanation:

1) Content of C:

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

Mass of C in 1.813 mg of CO₂

Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

Number of moles of C

number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

2) Content of H

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

Mass of H in 0.4639 mg of H₂O

Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

Number of moles of H

number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

3) Content of N

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

Mass of N in 0.2885 mg of N₂ is 0.2885 mg

Number of moles of N

number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

4) Content of O

The mass of O is calculated by difference:

Mass of O = mass of sample - mass of C - mass of H - mass of N

Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

Mass of O = 0.1648 mg

Moles of O = 0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

5) Ratios

Divide every number of mililmoles by the smallest number of milimoles:

C: 0.041195 / 0.01030 = 4
H: 0.051501 / 0.01030 = 5
N: 0.020597 / 0.01030 = 2

O: 0.01030 / 0.01030 = 1

C: 4
H: 5
N: 2
O: 1

6) Empirical formula:

C₄H₅N₂O₁

7) Calculate the approximate mass of the empirical formula:

4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 = 97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0

3 years ago

How many moles of KNO3 are in 500 mL of 2.0 M KNO3? ___ mol KNO3