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xz_007 [3.2K]
3 years ago
9

Which of the following is a solution?

Chemistry
1 answer:
Inessa05 [86]3 years ago
4 0
I think the answer would be B) Salt water because the other options wouldn't make much sense.
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Problem page what kind of intermolecular forces act between a dichlorine monoxide molecule and a hydrogen bromide molecule?
marta [7]
Dichlorine monoxide has the same structure like that of water. So, this is a polar molecule. For polar molecules, the dominant intermolecular force would be dipole-dipole forces. For HBr, there is a force between two oppositely charged ions, H⁺ and Br⁻. So, the dominant intermolecular force is electrostatic attraction.
5 0
3 years ago
Read 2 more answers
When a car uses a gallon of gas, how much carbon dioxide is emitted into the environment? I give branlyest
sineoko [7]

Answer:

19.5

Please tell me if wrong.

7 0
3 years ago
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Consider an ice cube and a hot radiator. Which has the higher thermal energy?
Law Incorporation [45]
The answer to your question is the radiator
8 0
2 years ago
What do you put in the calculator to get the molar mass of Mg3P2
olga55 [171]

Input the atomic masses of Mg and P to give 134.84g/mol

Explanation:

The molar mass of a substance (atom or molecule or compound) is the mass in grams of one mole of the substance:

When dealing with an element the molar mass is the relative atomic mass expressed as g/mol.

For compounds, you add the atomic masses of the component atoms and you sum up.

You simply input the atomic mass of 3 atoms of Mg and 2 atoms of P

Atomic mass of Mg = 24.3g/mol

                             P = 30.97g/mole

Molar mass of Mg₃P₂ = 3(24.3) + 2(30.97) = 134.84g/mol

learn more:

Molar mass brainly.com/question/2861244

#learnwithbrainly

6 0
3 years ago
The half life for the decay of carbon-14 is 5.73 x 10 years. Suppose the activity due to the radioactive decay of the carbon-14
Elena-2011 [213]

Answer:

Age of the atifact is 4.2\times 10^{2} years

Explanation:

  • For first -order radioactive decay- A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}
  • A_{t} represents activity of radioactive nuclide after t time, A_{0} represents initial activity of radioactive nuclide and t_{\frac{1}{2}} represents half-life
  • Here, A_{t}=19Bq, A_{0}=20Bq and t_{\frac{1}{2}}=5.73\times 10^{3}years

Plug-in all the given values in the above equation-

19=20\times (\frac{1}{2})^{\frac{t}{5.73\times 10^{3}}}

or, t=4.2\times 10^{2}

So, age of the atifact is 4.2\times 10^{2} years

6 0
3 years ago
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