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IrinaK [193]
3 years ago
15

How many grams do 1.5 x 1026 atoms of magnesium weigh?

Chemistry
1 answer:
LUCKY_DIMON [66]3 years ago
4 0

Answer:

0.3

Explanation:

if I'm wrong don't mark it against me

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When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas write the balanved equati
babunello [35]

Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}.

Explanation:

The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.

Now, in terms of chemical formulae this reaction equation will be as follows.

MnO_{2} + HCl \rightarrow H_{2}O + MnCl_{2} + Cl_{2}

Here, number of atoms on reactant side are as follows.

  • Mn = 1
  • O = 2
  • H = 1
  • Cl = 1

Number of atoms on product side are as follows.

  • Mn = 1
  • O = 1
  • H = 2
  • Cl = 4

To balance this equation, multiply HCl by 4 on reactant side and multiply H_{2}O by 2 on product side. Therefore, the equation can be rewritten as follows.

MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}

Hence, number of atoms on reactant side are as follows.

  • Mn = 1
  • O = 2
  • H = 4
  • Cl = 4

Number of atoms on product side are as follows.

  • Mn = 1
  • O = 2
  • H = 4
  • Cl = 4

Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.

Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}.

6 0
3 years ago
What substance is oxidized in the reaction Mg + Cu2+ — Mg2+ + Cu?
alexgriva [62]

Answer:

Mg

Explanation:

Mg + Cu2+ — Mg2+ + Cu

Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

<em>Consider the following reactions. </em>

4KI + 2CuCl₂  →   2CuI  + I₂  + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O   →  CO₂ + H₂

the oxidation state of carbon is +2 on reactant  side and on product side it becomes  +4 so carbon get oxidized.

Na₂CO₃ + H₃PO₄  →  Na₂HPO₄ + CO₂ + H₂O

The oxidation state of carbon on reactant side is +4. while on product side is  also +4 so it neither oxidized nor reduced.

H₂S + 2NaOH → Na₂S + 2H₂O

The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.

Oxidizing agents:

Oxidizing agents oxidize the other elements and itself gets reduced.

Reducing agents:

Reducing agents reduced the other element are it self gets oxidized.

8 0
3 years ago
Which properties cause chemical reactions?
GREYUIT [131]
D. Chemical and Physical Properties
3 0
3 years ago
Read 2 more answers
Determine the number of liters in a 355 milliliter can of soda.<br> Ans: 0.355 L
Drupady [299]
0.355

A cubic inch is a unit of volume. It is the size of a cube that is 1 inch on a side. It is approximately 1.1 tablespoons, or about 16.4 cubic centimeters. A liter, or litre, is a unit of volume in the metric system.
...
8 0
3 years ago
Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

8 0
2 years ago
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