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emmasim [6.3K]
3 years ago
5

An atom has a negative charge. Which of the following must have occurred?

Chemistry
1 answer:
galina1969 [7]3 years ago
5 0
It should be A)It lost a neutron.
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What is the vapor pressure of a solution that contains 70.0 mL of ether and 30.0 mL of benzene?
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Answer:

it is 50

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6 0
3 years ago
CHEMISTRY OuO Help please
jeka94

Answer:

This is google's answer for the last question

Explanation:

The kinetic energy increases as the particles move faster. The potential energy increases as the particles move farther apart. How are thermal energy and temperature related? When the temperature of an object increases, the average kinetic energy of its particles increases.

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3 years ago
Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such
8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

  • 1 A = 1 C/s
  • The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
  • 1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
  • The molar mass of silver is 107.87 g/mol

The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g

5 0
3 years ago
Please help with 2 chem questions thank you!
xeze [42]
The correct answers are A and C.
8 0
3 years ago
Read 2 more answers
In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.
olga nikolaevna [1]

Answer:

a. The second run will be faster.

d. The second run has twice the surface area.

Explanation:

The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.

A=\pi ^{1/3} (6V)^{2/3}

The area of the 10.0 cm³-sphere is:

A=\pi ^{1/3} (6.10.0)^{2/3}=22.4cm^{2}

The area of each 1.25 cm³-sphere is:

A=\pi ^{1/3} (6. 1.25)^{2/3}=5.61cm^{2}

The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²

The ratio of  8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00

Since the surface area is doubled, the second run will be faster.

6 0
3 years ago
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