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Salsk061 [2.6K]
2 years ago
11

What is chemistry? ​

Chemistry
1 answer:
kirill [66]2 years ago
3 0

Answer:the branch of science which examines the material of the universe and changes that these materials undergo

Chemistry deals with composition,structure,properties of matter and changes that these materials undergo

You might be interested in
Jumping on a cemented floor is more pain full than a sandy flour,why​
Anvisha [2.4K]

Answer:

Is soft

Explanation:

because concrete is hard ash so t hink flour would be safer

5 0
3 years ago
The molarity of a solution prepared by dissolving 4.11 g of NaI in enough water to prepare 312 mL of solution is
Dimas [21]

Answer:

The correct answer is 8.79 × 10⁻² M.

Explanation:

Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,

No. of moles of NaI = Weight of NaI/ Molecular mass

= 4.11 / 149.89

= 0.027420

The vol. of the solution given is 312 ml or 0.312 L

The molarity can be determined by using the formula,

Molarity = No. of moles/ Volume of the solution in L

= 0.027420/0.312

= 0.0879 M or 8.79 × 10⁻² M

6 0
2 years ago
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
2 years ago
What is the end result of a nuclear fusion reaction?
julia-pushkina [17]

Answer:A.) molecule of atoms forms

Explanation:

7 0
3 years ago
Read 2 more answers
Convert 14.72 kg to ____ mg
navik [9.2K]

Answer:

14720000

Explanation:

1 kg = 1000000 mg

14.72 kg = 14.72 x 1000000

=14720000

Please Mark me brainliest

6 0
2 years ago
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