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swat32
3 years ago
11

Will bromine water and magnesium react

Chemistry
1 answer:
bazaltina [42]3 years ago
7 0
No. Magnesium, and Bromine are a chemical compound when put together.
Neither Bromine, nor Magnesium react with any sort of water.
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Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

6 0
2 years ago
B) How many kilograms of carbon dioxide are formed when 24.42 g of iron is<br> produced?
charle [14.2K]

Answer:

0.0289 kg of CO2 will be formed

Explanation:

Step 1: Data given

Mass of iron produced = 24.42 grams

Atomic mass iron = 55.845 g/mol

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

Fe2O3 + 3CO → 2Fe + 3CO2

Step 3: Calculate moles iron

Moles iron = mass iron / molar mass iron

Moles iron = 24.42 grams / 55.845 g/mol

Moles iron = 0.437 moles

Step 4: Calculate moles CO2

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 0.437 moles Fe we'll have 3/2 * 0.437 = 0.6555 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.6555 moles * 44.01 g/mol

Mass CO2 = 28.85 grams = 0.0289 kg

0.0289 kg of CO2 will be formed

5 0
3 years ago
Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338
Elden [556K]

<u>Answer:</u> The concentration of chloride ions in the solution obtained is 0.674 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

  • <u>For KCl:</u>

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

  • <u>For magnesium chloride:</u>

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = (2\times 0.224)=0.448mol

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

5 0
2 years ago
Complete by describing what the change of the state of the matter will be for the process to occur
Brilliant_brown [7]

Answer:

Freezing: Liquid turns into solid. Evaporation: Liquid turns into gas. Condensation: Gas turns into liquid. Sublimation: solid into gas. Deposition: Gas into solid.

Explanation:

4 0
2 years ago
Post laboratory reflections 1 write equation [2] in the net ionic form 2. describe what happens when strong acid acts on limesto
oee [108]
It deteriorates. Definitely more than when you started.

4 0
2 years ago
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