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3 years ago

The reactants of two chemical reactions are shown.

Chemistry

2 answers:

Ludmilka [50]3 years ago

6 0

Answer:

Explanation:

The first step is to write the whole reactions:

A) $HCl~+~NaOH~->~H_2O~+~NaCl$

B) $Cu~+~AgNO_3~->~Cu(NO_3)_2~+~Ag$

In A) reaction we will have an acid-base reaction (double replacement reaction). In this reaction, the oxidation number is the same for all atoms on both sides of the reaction.

In B) the oxidation number of Cu will change from zero to +2 and the oxidation number of Ag will change from +1 to zero. So, the Cu will be oxidized and the Ag will reduce. So, if we have a redox reaction for B) the charge must be equal on both sides.

kozerog [31]3 years ago

3 0

Answer:

it is A

Explanation:

because the forward reation will give salt and water why the backward reaction will give base and acid

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if an element has an atomic mass of 127 and an atomic number of 53, how many neutrons does it have? 74 127 53 75

nadya68 [22]

74 is the answer u find it by subtracting the protons(atomic number) from atomic mas, so basically... 127-53 =74 neutrons

7 0

3 years ago

You are provided with a stock solution with a concentration of 1.0x10-5 M. You will be using this to make two standard solutions

artcher [175]

Answer:

1. V₁ = 2.0 mL

2. V₁ = 2.5 mL

Explanation:

You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.

To calculate the volume required (V₁) in each dilution we will use the dilution rule.

C₁ . V₁ = C₂ . V₂

where,

C are the concentrations

V are the volumes

1 refers to the initial state

2 refers to the final state

1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.

C₁ . V₁ = C₂ . V₂

(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL

V₁ = 2.0 mL

2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.

C₁ . V₁ = C₂ . V₂

(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL

V₁ = 2.5 mL

8 0

3 years ago

Which atoms in the cfc molecule can destroy thousands of ozone molecules in the upper atmosphere?

zalisa [80]

Fluorine and chlorine

6 0

3 years ago

Read 2 more answers

What happens to the volume of a gas in a closed container if the temperature increases, but the pressure remains the same? Why?

labwork [276]

Answer:

Volume will goes to increase.

Explanation:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

So when the temperature goes to increase the volume of gas also increase. Higher temperature increase the kinetic energy and molecules move randomly every where in given space so volume increase.

Now we will put the suppose values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 4.5 L × 348 K / 298 k

V₂ = 1566 L.K / 298 K

V₂ = 5.3 L

Hence prove that volume increase by increasing the temperature.

3 0

3 years ago

Combustion reactions are exothermic. The heat of reaction for the combustion of 2-methylheptane, C8H18, is 1.306×103 kcal/mol. W

WARRIOR [948]

Answer:

11.45kcal/g

2.612 × 10³ kcal

Explanation:

When a compound burns (combustion) it produces carbon dioxide and water. The combustion of 2-methylheptane can be represented by the following balanced equation:

2 C₈H₁₈ + 25 O₂ ⇄ 16 CO₂ + 18 H₂O

It releases 1.306 × 10³ kcal every 1 mol of C₈H₁₈ that is burned.

What is the heat of combustion for 2-methylheptane in kcal/gram?

We know that the molar mass of C₈H₁₈ is 114.0g/mol. Then, using proportions:

$\frac{1.306 \times 10^{3}Kcal}{1mol} .\frac{1mol}{114.0g} =11.45kcal/g$

How much heat will be given off if molar quantities of 2-methylheptane react according to the following equation? 2 C₈H₁₈ + 25 O₂ ⇄ 16 CO₂ + 18 H₂O

In this equation we have 2 moles of C₈H₁₈. So,

$2mol \times\frac{1.306 \times 10^{3}kcal }{1mol} =2.612\times 10^{3}kcal$

3 0

3 years ago