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3 years ago

A sample of sugar (C12H22O11) contains

Chemistry

1 answer:

Lady_Fox [76]3 years ago

5 0

4 moles of sugar.

Explanation:

A mole is defined as the amount of a substance contained in Avogadro's number of particles 6.02 x 10²³.

1 mole of substance = 6.02 x 10²³. molecules

Given that;

the sample of sugar contains 1.505 x 10²³.molecules

The number of moles in this amount of sugar is 4 moles

Learn more:

Number of moles brainly.com/question/13064292

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Which of the following is a unit of mass? A) A meter B) A liter C) a kelvin D) a kilogram

lora16 [44]

A kilogram is a unit of mass. A meter is a unit of measurement, a Liter is volume and a kelvin is temperature

6 0

3 years ago

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27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams

GarryVolchara [31]

Answer: The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aluminium nitrite:

Given mass of aluminium nitrite = 27.4 g

Molar mass of aluminium nitrite = 41 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol$

For ammonium chloride:

Given mass of ammonium chloride = 169.9 g

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol$

The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:

$Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O$

By Stoichiometry of the reaction:

1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride

So, 0.668 moles of aluminium nitrite will react with = $\frac{3}{1}\times 0.668=2.004mol$ of ammonium chloride.

As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.

Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles

Calculating the mass of ammonium chloride by using equation 1, we get:

Excess moles of ammonium chloride = 1.172 moles

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

$1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g$

Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

3 0

3 years ago

In a test of an automobile engine 1.00 L of octane (702 g) is burned, but only 1.84 kg of carbon dioxide is produced. What is th

Reptile [31]

Answer:

The % yield of CO2 is 85.05 %

Explanation:

Step 1: Data given

Mass of octane = 702 grams

Molar mass octane = 114.23 g/mol

Mass CO2 =1.84 kg = 1840 grams

Molar mass of CO2

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles of octane

Moles octane = mass octane / molar mass octane

Moles octane = 702.0 grams / 114.23 g/mol

Moles octane = 6.145 moles

Step 4: Calculate moles of CO2

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

For 6.145 moles octane we'll have 8*6.145 moles =49.16 moles

Step 5: Calculate mass of CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 49.16 moles * 44.01 g/mol

Mass CO2 = 2163.5 grams

Step 6: Calculate % yield of carbon dioxide

% yield = (actual yield / theoretical yield)*100%

% yield = (1840/2163.5)*100%

% yield = 85.05 %

The % yield of CO2 is 85.05 %

4 0

3 years ago

What is the value of the equilibrium constant for this redox reaction?

aivan3 [116]

Correct answer: Option D, K = 5.04 × 10^52

Reason:
We know that,
Ecell = $\frac{0.0592}{n}log(K)$ ,
where n = number of electrons = 2 (in present case)
K = equilibrium constant.

Also, Ecell = +1.56 v

Therefore, 1.56 = $\frac{0.0592}{2}log(K)$
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52

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3 years ago

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A rectangle solid of unknown density is 5 m long two meters high and 4 meters wide. The mass of this solid is 300 grams. Find th

aksik [14]

D= M/V so, D=300/40 wich is 7.5 so D= 7.5g/m3

6 0

2 years ago