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3 years ago

Which of the following changes

Chemistry

1 answer:

Alisiya [41]3 years ago

5 0

Exothermic reactions

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Chlorine + sulfur dioxide + water = hydrochloric acid + sulfuric acid

scZoUnD [109]

The given example is a chemical reaction.

The contents (separated as reactants and products) :

$\begin{tabular}{c | l}Reactants & Products \\\cline{1-2}Chlorine & Sulfuric Acid \\Water & Hydrochloric Acid \\Sulfur Dioxide & \\\end{tabular}$

The written reaction is :

$\boxed {Cl + SO_{2} + H_{2}O \implies HCl + H_{2}SO{4}}$

I hope it helped you solve the problem.

Good luck on your studies!

7 0

1 year ago

The Lewis dot notation for two atoms is shown.

Iteru [2.4K]

Answer:

Mg donates two electrons to O

Explanation:

Lewis dot notation uses dots and crosses to represent valence electrons on atoms.

Magnesium is a metal and would donate or lose electrons during bonding.

Oxygen is a non metal and would gain electrons during bonding.

The correct option is;

Mg donates two electrons to O

5 0

2 years ago

Which of the following lists elements from lowest to highest atomic number

vitfil [10]

Lowest is Hydrogen highest is Beryllium
-HOPE THIS HELPED

8 0

3 years ago

Why is the formation of fructose-1,6-bisphosphate a step in which control is likely to be exercised in the glycolytic pathway

Yakvenalex [24]

The answer is/6-(1-2)

Explain why it is not that because it shown

5 0

2 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

Best regards.

8 0

3 years ago