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Which products would form if chlorine gas was bubbled through a solution of sodium bromide?

777dan777 [17]

Answer:

i think d

Explanation:

4 0

3 years ago

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How many moles of glucose, C H O , can be "burned" biologically when 18.2 mol of oxygen is available? C H O (s) + 6O (g) --->

Feliz [49]

Answer:
547.7 g of C₆H₁₂O₆

Solution:
The balance chemical equation is as follow,

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

According to equation,

6 moles of O₂ burns = 180.56 g of C₆H₁₂O₆
So,
18.2 moles of O₂ will burn = X g of C₆H₁₂O₆

Solving for X,
X = (18.2 mol × 180.56 g) ÷ 6 mol

X = 547.7 g of C₆H₁₂O₆

5 0

3 years ago

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Is relative dating an effective way to determine the age of rocks?

JulsSmile [24]

Answer: Read Explanation

Explanation: Index fossils are useful because they tell the relative ages of the rock layers in which they occur. Geologists use particular types of organisms, such as trilobites, as index fossils.

7 0

3 years ago

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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

How many molecules of Oxygen are consumed if 24 molecules of Carbon Dioxide are produced?

Nat2105 [25]

Answer:

Explanation:

C + O2 → CO2

Mole of C = 24 g/(12 g/mole)

Mole of C = 2 mole

Mole of molecular O2 = 74 g/(32 g/mole)

Mole of molecular O2 = 2.3125 mole

Since mole of C < mole of O2, then C being the limiting reagent.

From the reaction, it shows that mole ratio between C and O2 = 1 : 1.

So, 2 moles of C will stoichiometrically react with 2 moles of O2 to generate 2 moles of CO2.

Avogadro's law states that :"equal volumes of all gases, at the same temperature and pressure, have the same number of molecules i.e. 6.02 x 10^23 molecules/mole.

Therefore, 2 moles of CO2 contain 2 moles x 6.02 x 10^23 molecules/mole = 1.204 x 10^24 molecules of CO2 is formed.

6 0

3 years ago