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2 years ago

How many grams of Kr are in 2.00 mol of Kr? (4)

Chemistry

2 answers:

Lina20 [59]2 years ago

8 0

Answer:

1 mole is equal to 1 moles Krypton, or 83.798 grams.

Explanation:

andrew11 [14]2 years ago

3 0

ANSWER: 168 grams of Kr
This is the rounded answer, but the full answer is 168.7 grams (I’m not sure if you need to use significant figures or not).

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A student tries to make diphosphorus trioxide using the reaction below. He adds exactly 20.00 grams of Phosphorus to an excess o

scoray [572]

Answer:

Mass = 87.952 g

Explanation:

Given data:

Mass of phosphorus = 20.0 g

Theoretical yield of P₂O₃ = ?

Solution:

Chemical equation:

P₄ + 3O₂ → 2P₂O₃

Number of moles of phosphorus:

Number of moles = mass/ molar mass

Number of moles = 20.0 g / 123.9 g/mol

Number of moles = 0.2 mol

Now we will compare the moles of phosphorus with P₂O₃.

P₄ : P₂O₃

1 : 2

0.2 : 2/1×0.2 = 0.4 mol

Theoretical yield:

Mass = number of moles × molar mass

Mass = 0.4 mol × 219.88 g/mol

Mass = 87.952 g

7 0

3 years ago

What is it called when a consume wastes and dead organisms??????

8_murik_8 [283]

Decomposers and a few insects.

8 0

2 years ago

How many neutrons does an element have if its atomic number is 41 and its mass number is 170?

BARSIC [14]

atomic number = protons.

protons+neutrons=atomic mass.

170 (mass) - 41 (proton/number) = 129

that should be ur answer. I hope this helps!

7 0

3 years ago

If you start with 0.05 M O3 and 0.01 M NO and the reaction reaches completion in 16 seconds, what is the initial rate of this re

makvit [3.9K]

Since we are basing the rate with respect to O3, therefore the rate would simply be the concentration divided by the total reaction time. That is:

rate of reaction = 0.05 M / 16 seconds

rate of reaction = 3.125 x 10^-3 M / second

rate of reaction = 0.003125 M / second

6 0

3 years ago

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