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3 years ago

Could someone pls help me if you know it!!

Chemistry

1 answer:

klio [65]3 years ago

3 0

Answer:

Beautiftul, begging under F

honest, beginning in thrid line under A

hopeful, fifth line under F

Explanation:

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For a reversible reaction, what would a large equilibrium constant indicate?

siniylev [52]

(D) At equilibrium, the concentration of the products will be much higher than the concentration of the reactants.

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4 years ago

How do you know when a redox equation is balanced?

professor190 [17]

Like to check? Just add up the charges on both sides and see if they equal each other:)

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3 years ago

A volume of 60.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What

mihalych1998 [28]

Answer:

The molarity of KOH = 0.86M

Explanation:

From the equation of reaction

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

nA= 1, nB=2

From the question

CA= 1.5M, VA= 17.2ml, VB= 60ml, CB= ?

Applying (CAVA)/(CBVB) = nA/nB

(1.5×17.2)/(CB×60) = 1/2

Simplify

CB= (1.5×17.2×2)/(60×1)

CB = 0.86M

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3 years ago

What is the process called when rocks are broken down into smaller pieces over time by the environment

Arte-miy333 [17]

Answer:

erosion

Explanation:

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3 years ago

Read 2 more answers

What is the theoretical yield in grams of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuS

vfiekz [6]

Answer : The theoretical yield of CuS is, 7.26 grams.

Solution : Given,

Mass of $Na_2S$ = 15.5 g

Mass of $CuSO_4$ = 12.1 g

Molar mass of $Na_2S$ = 78 g/mole

Molar mass of $CuSO_4$ = 160 g/mole

Molar mass of CuS = 96 g/mole

First we have to calculate the moles of $Na_2S$ and $CuSO_4$ .

$\text{ Moles of }Na_2S=\frac{\text{ Mass of }Na_2S}{\text{ Molar mass of }Na_2S}=\frac{15.5g}{78g/mole}=0.199moles$

$\text{ Moles of }CuSO_4=\frac{\text{ Mass of }CuSO_4}{\text{ Molar mass of }CuSO_4}=\frac{12.1g}{160g/mole}=0.0756moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS$

From the balanced reaction we conclude that

As, 1 mole of $CuSO_4$ react with 1 mole of $Na_2S$

So, 0.0756 moles of $CuSO_4$ react with 0.0756 moles of $Na_2S$

From this we conclude that, $Na_2S$ is an excess reagent because the given moles are greater than the required moles and $CuSO_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CuS$

From the reaction, we conclude that

As, 1 mole of $CuSO_4$ react to give 1 mole of $CuS$

So, 0.0756 moles of $CuSO_4$ react to give 0.0756 moles of $CuS$

Now we have to calculate the mass of $CuS$

$\text{ Mass of }CuS=\text{ Moles of }CuS\times \text{ Molar mass of }CuS$

$\text{ Mass of }CuS=(0.0756moles)\times (96g/mole)=7.26g$

Thus, the theoretical yield of CuS is, 7.26 grams.

8 0

3 years ago