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ddd [48]
3 years ago
8

When 35.0 mL of 0.400 M hydrobromic acid and 35.0 mL of 0.400 M ammonia are combined, the pH of the resulting solution will be :

..... B A. Less than 7 B. Equal to 7 C. Greater than 7
Chemistry
1 answer:
Kay [80]3 years ago
3 0

Answer:

B. Equal to 7.

Explanation:

Hydrobromic acid is a strong acid that decreases pH and ammonia is a strong base that increases pH.

As the initial pH of water is 7,0 the addition of 35.0mL of 0.400M HBr will produce a pH less than 7,0. But, the same effect of decreasing pH is reverted for the addition of 35.0mL of 0.400M HNO3.

That means the net effect of the two addition is to have a pH:

B. Equal to 7.

I hope it helps!

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Mice21 [21]

Answer:A

Explanation:

7 0
3 years ago
How do solve for #13?What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?
exis [7]

What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?

The change in Boiling Point of water can be calculated using this formula:

ΔTb = i * Kb * m

Where i is the van't hoff factor (the number of particles or ions), the kb is a constant (boiling point elevation constant) and m is the molality of the solution.

The kb for water is always 0.515 °C/m. Kb = 0.515 °C/m

The value for i in this case is 1. Since sucrose is a covalent compound and it doesn't dissociate into ions. i = 1

The molal concentration of the solution can be found using this formula:

molality = moles of sucrose/kg of water

molality = 1.000 mol / 1.000 kg of water

molality = 1 m

Now that we know all the values, we can use the formula to find the change in the boiling point of water:

ΔTb = i * Kb * m

ΔTb = 1 * 0.515 °C/m * 1 m

ΔTb = 0.515 °C

Finally, we are asked for the boiling point of the solution, not the change. The boiling point of water at atmospheric pressure is 100.00 °C. If the boiling point rises 0.515 °C when we prepare the solution. The boiling point of the solution is:

Boiling point solution = Boiling point of water + ΔTb

Boiling point solution = 100.000 °C + 0.515 °C

Boiling point solution = 100.515 °C

Answer: The boiling point of the solution is 100.515 °C.

8 0
1 year ago
The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p
Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

    Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g  

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}

(ii) Solve the relationship

 Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

    56x + 18(7.5 - x) = 267.3

      56x + 135 - 18x = 267.3

                        38x = 132.3

                            x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

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Answer:

it is b

Explanation:

A drainage basin is any area of land where precipitation collects and drains off into a common outlet, such as into a river, bay, or other body of water

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