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3 years ago

Worth 13 points plzzz helpp

Chemistry

2 answers:

Annette [7]3 years ago

8 0

Answer: Another K12 student, fun! Also, you may want to delete this because you just displayed your full name..

serious [3.7K]3 years ago

5 0

The fourth one is correct yw

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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago

in a mixture of 1.90 mol of gas, 0.85 mol are nitrogen (n2) molecules. what is the mole fraction of n2 in this mixture?

Nostrana [21]

0.447 is the mole fraction of Nitrogen in this mixture.

mole fraction of nitrogen= moles of nitrogen/total moles

mole fraction of nitrogen=0.85/1.90

mole fraction of nitrogen=0.447

The product of the moles of a component and the total moles of the solution yields a mole fraction, which is a unit of concentration measurement. Because it is a ratio, mole fraction is a unitless statement. The sum of the components of the mole fraction of a solution is one. In a mixture of 1 mol benzene, 2 mol carbon tetrachloride, and 7 mol acetone, the mole fraction of the acetone is 0.7. This is computed by dividing the sum of the moles of acetone in the solution by the total number of moles of the solution's constituents:

To know more about mole fraction visit : brainly.com/question/8076655

#SPJ4

4 0

1 year ago

If you were given two transparent liquids how would you determine which is pure and which is salt water

Tpy6a [65]

Based off the salt levels in each water.

8 0

3 years ago

If the CaCO3 weighed 983 g and the CaO weighed 551 g, how many grams of CO2 were formed in the reaction?

stira [4]

Answer:

The answer to your question is 432 g of CO₂

Explanation:

Data

CaCO₃ = 983 g

CaO = 551 g

CO₂ = ?

Balanced reaction

CaCO₃ (s) ⇒ CaO (s) + CO₂ (g)

This reaction is balanced, to solve this problem just remember the Lavoisier Law of conservation of mass that states that the mass of the reactants is equal to the mass of the products.

Mass of reactants = Mass of products

Mass of CaCO₃ = Mass of CaO + Mass of CO₂

Solve for CO₂

Mass of CO₂ = Mass of CaCO₃ - Mass of CaO

Mass of CO₂ = 983 g - 551 g

Simplification

Mass of CO₂ = 432 g

5 0

2 years ago

You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.200 M sodium benzoate

bekas [8.4K]

Answer : The volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

Explanation :

Let the volume of sodium benzoate (salt) be, x

So, the volume of benzoic acid (acid) will be, (100 - x)

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$4.00=4.20+\log \left(\frac{(\frac{0.200x}{100})}{(\frac{0.100(100-x)}{100})}\right)$

x = 29.0

The volume of sodium benzoate = x = 29.0 mL

The volume of benzoic acid (acid) = (100 - x) = (100 - 29.0) = 71 mL

Thus, the volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

5 0

3 years ago