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If a compound is found to contain 3.622 % carbon and 96.38 % bromine by weight. The molecular formula for the compound is CBr4.

First, get the empirical formula in order to calculate the molecular formula of the chemical. The empirical formula shows the atoms of a compound in their most straightforward whole number ratio.

Suppose 100 grams of the substance. To determine the mass of carbon and bromine in the compound using the provided percentages.

Mass of C = 3.622% of 100g

= 3.622g

Mass of Br = 96.38% of 100g

= 96.38g

The next step is to determine the atomic masses of carbon and bromine in order to determine the number of moles for each.

Atomic mass of carbon = 12.01 g/mol

Atomic mass of bromine = 79.90 g/mol

Moles of C = (mass of carbon) / (atomic mass of carbon)

= 3.622g / 12.01 g/mol

= 0.3016 mol

Moles of Br = (mass of bromine) / (atomic mass of bromine)

= 96.38g / 79.90 g/mol

= 1.205 mol

Divide the moles of each element by the fewest number of moles obtained, in this case the moles of carbon, to arrive at the empirical formula.

Empirical formula ratio:

C: (0.3016 mol) / (0.3016 mol)

= 1

Br: (1.205 mol) / (0.3016 mol)

= 4

The empirical formula for the compound is C₁Br4.

To determine the molecular formula, it is required to know the molecular weight of the compound. The molecular weight is  331.61 g/mol.

To find the number of empirical formula units in the molecular formula, divide the molecular weight by the empirical formula weight.

Empirical formula weight:

C = 12.01 g/mol × 1

= 12.01 g/mol

Br= 79.90 g/mol × 4

= 319.60 g/mol

Empirical formula weight = 12.01 + 319.60

= 331.61 g/mol

Now find the number of empirical formula units in the molecular formula:

Number of empirical formula units

= (molecular weight) ÷ (empirical formula weight)

Number of empirical formula units

= 331.61 g/mol / 331.61 g/mol

= 1

The number of empirical formula units is 1, the empirical formula C₁Br4 is would be  the molecular formula for this compound.

Thus, the molecular formula for the compound is CBr₄.

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Balance the following chemical equations:

1. N2 + 3 H2 → 2 NH3

Ex: Synthesis reaction

2. 2 KClO3 → 2 KCl + 3 O2

Single Replacement reaction

3. 2 NaF + ZnCl2 → ZnF2 + 2 NaCl

Decomposition reaction

4. 2 AlBr3 + 3 Ca(OH)2 → Al2(OH)6 + 6 CaBr2

Double Replacement reaction

5. 2 H2 + O2 → 2 H2O

Combustion reaction

6. 2 AgNO3 + MgCl2 → 2 AgCl + Mg(NO3)2

Synthesis reaction

7. 2 Al + 6 HCl → 2 AlCl3 + 3 H2

Decomposition reaction

8. C3H8 + 5 O2 → 3 CO2 + 4 H2O

Combustion reaction

9. 2 FeCl3 + 6 NaOH → Fe2O3 + 6 NaCl + 3 H2O

Double Replacement reaction

10. 4 P + 5 O2 → 2 P2O5

Synthesis reaction

11. 2 Na + 2 H2O → 2 NaOH + H2

Single Replacement reaction

12. 2 Ag2O → 4 Ag + O2

Decomposition reaction

13. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

Combustion reaction

14. 2 KBr + MgCl2 → 2 KCl + MgBr2

Double Replacement reaction

15. 2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O

Double Replacement reaction

16. C5H12 + 8 O2 → 5 CO2 + 6 H2O

Combustion reaction

17. 4 Al + 3 O2 → 2 Al2O3

Synthesis reaction

18. Fe2O3 + 2 Al → 2 Fe + Al2O3

Single Replacement reaction

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Answer:

85.99 grams of AgCl will be formed.

Explanation:

To determine the grams of AgCl formed in the reaction, we need to consider the stoichiometry of the reaction between silver nitrate (AgNO3) and barium chloride (BaCl2):

AgNO3 + BaCl2 -> AgCl + Ba(NO3)2

The balanced equation shows that the molar ratio between AgNO3 and AgCl is 1:1. This means that 1 mole of AgNO3 produces 1 mole of AgCl.

Given:

Volume of BaCl2 solution = 1500 ml = 1.5 L

Molarity of BaCl2 solution = 0.400 M

First, we need to calculate the number of moles of BaCl2 present in the solution:

moles of BaCl2 = volume of BaCl2 solution * molarity of BaCl2 solution

= 1.5 L * 0.400 M

= 0.600 moles

Since the molar ratio between BaCl2 and AgNO3 is 1:1, the number of moles of AgNO3 needed for complete reaction is also 0.600 moles.

Now, using the molar mass of AgCl, which is 143.32 g/mol, we can calculate the grams of AgCl formed:

grams of AgCl = moles of AgNO3 * molar mass of AgCl

= 0.600 moles * 143.32 g/mol

= 85.99 grams

Therefore, by adding enough AgNO3 to react fully with the 1500 ml of 0.400 M BaCl2 solution, approximately 85.99 grams of AgCl will be formed.

Answer:

2.282 atm

P1V1/T1 = P2V2/T2

2.70atm / (50+273) = X/ 273

make x subject of formula

:. X = 2.28 atm

or 2.28 * 1.01 *10⁵ N/m²

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The volume of the gas at standard temperature and pressure conditions is approximately 12.77 liters.

To find the volume of the gas at STP, we can use the combined gas law:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

Note that: at STP (Standard Temperature and Pressure) is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm.

Given that:

P₁ = initial pressure = 4.71 atm

V₁ = initial volume = 2.99 L

T₁ = initial temperature = 28.10 °C = ( 28.10 + 273.15 ) = 301.25 K

P₂ = final pressure (STP pressure ) = 1 atm

T₂ = final temperature (STP temperature)  = 0°C = 273.15 K

V₂ = final volume = ?

Substituting the given values into the formula:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\\\frac{4.71\ *\ 2.99 }{301.25} = \frac{1\ *\ V_2}{273.15 }\\\\V_2 = 12.77\ L[/tex]

Therefore, the final volume is 12.77 litres.

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The molarity of the solution, if 44g of [tex]CaCl_{2}[/tex] is dissolved in 95 ml of water is 4.1733 M

To calculate the molarity (M) of a solution, we use the formula:

Molarity (M) = moles of solute/volume of solution in liters

As per the question:

Mass of [tex]CaCl_{2}[/tex] = 44 g

Volume of water = 95 mL = 0.095 L

To find molarity, we need to determine the number of moles of [tex]CaCl_{2}[/tex] by dividing the given mass by its molar mass.

Molar mass of [tex]CaCl_{2}[/tex] = 40.08 g/mol (for [tex]Ca[/tex]) + (2 × 35.45 g/mol) (for [tex]Cl[/tex])

Molar mass of [tex]CaCl_{2}[/tex] = 110.98 g/mol

Number of moles of [tex]CaCl_{2}[/tex] = Mass of [tex]CaCl_{2}[/tex] / Molar mass of [tex]CaCl_{2}[/tex]

Number of moles of [tex]CaCl_{2}[/tex] = 44 g / 110.98 g/mol

Number of moles of [tex]CaCl_{2}[/tex] ≈ 0.3965 mol

Now, to calculate the molarity of the solution, we can use this formula:

Molarity (M) = moles of solute/volume of solution in liters

Molarity (M) = 0.3965 mol / 0.095 L

Molarity (M) ≈ 4.1733 M

Therefore, the molarity of the solution is approximately 4.1733 M when 44 g of [tex]CaCl_{2}[/tex] is dissolved in 95 mL of water.

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