Balance the following chemical reaction. ____NaOH+____H2SO4 → ___Na2SO4+___

Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions.

Norma-Jean [14]

$2 KHP(aq) + Ba(OH)_{2}(aq) ---> KP_{2}Ba (aq) + 2 H_{2}O (l)$ \

Moles of $Ba(OH)_{2}$ = $0.483 \frac{mol}{L} * 28.5 mL *\frac{1 L}{1000mL} = 0.0138 mol Ba(OH)_{2}$

Mass of KHP = $0.0138 mol Ba(OH)_{2} * \frac{2 mol KHP}{1 mol Ba(OH)_{2}} * \frac{204.22 g KHP}{1mol KHP}$

= 5.64 g KHP

6 0

4 years ago

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36.8 g of CuSO4(s) was added to water to prepare a 2.00 L solution. What is the

Yuki888 [10]

Answer:

Option D. 0.115 M

Explanation:

The following data were obtained from the question:

Mass of CuSO4 = 36.8 g

Volume of solution = 2 L

Molar mass of CuSO4 = 159.62 g/mol

Molarity of CuSO4 =..?

Next, we shall determine the number of mole in 36.8 g of CuSO4.

This can be obtained as shown below:

Mass of CuSO4 = 36.8 g

Molar mass of CuSO4 = 159.62 g/mol

Mole of CuSO4 =.?

Mole = mass /Molar mass

Mole of CuSO4 = 36.8 / 159.62

Mole of CuSO4 = 0.23 mole

Finally, we shall determine the molarity of the CuSO4 solution as illustrated below:

Mole of CuSO4 = 0.23 mole

Volume of solution = 2 L

Molarity of CuSO4 =..?

Molarity = mole /Volume

Molarity of CuSO4 = 0.23 / 2

Molarity of CuSO4 = 0.115 M

Therefore, the molarity of the CuSO4 solution is 0.115 M.

5 0

4 years ago

.What type of energy includes both kinetic and potential energy?

PSYCHO15rus [73]

Answer:

mechanical energy

Explanation:

6 0

3 years ago

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What volume (in liters, at 703 k and 2.04 atm) of chlorine gas is required to react with 3.39 g of p?

Natali5045456 [20]

The volume of chlorine required is 7.71 L.

The reaction between phosphorus and chlorine is:

2P + 5Cl₂→ 5PCl₅

Therefore, 2 moles of P requires 5 moles of chlorine to react with it.

Given mass of P =3.39 g

Molar mass of P=30.97 g/mol

No. of moles of P = given mass/ molar mass = 3.39 / 30.97 = 0.109 moles

2 moles of P requires 5 moles of chlorine

0.109 moles of P will require 0.109 x 5/2 = 0.2725 moles of chlorine

According to ideal gas equation

PV=nRT

2.04 x V = 0.2725 x 0.0821 x 703

V = 0.2725 x 0.0821 x 703 / 2.04

V = 7.71L

Learn more about ideal gas equation here:

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5 0

2 years ago

Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/

Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of $C_2H_4$ and $H_2$ = 25.0 atm

Temperature of $C_2H_4$ and $H_2$ = $250^oC=273+250=523K$

Volume of $C_2H_4$ = 1050 L per min

Volume of $H_2$ = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of $C_2H_6$ = 30 g/mole

First we have to calculate the moles of $C_2H_4$ and $H_2$ by using ideal gas equation.

For $C_2H_4$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=611.34moles$

For $H_2$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=902.46moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_2H_4+H_2\rightarrow C_2H_6$

From the balanced reaction we conclude that

As, 1 mole of $C_2H_4$ react with 1 mole of $H_2$

So, 611.34 mole of $C_2H_4$ react with 611.34 mole of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $C_2H_6$ .

As, 1 mole of $C_2H_4$ react to give 1 mole of $C_2H_6$

As, 611.34 mole of $C_2H_4$ react to give 611.34 mole of $C_2H_6$

Now we have to calculate the mass of $C_2H_6$ .

$\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6$

$\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g$

The theoretical yield of $C_2H_6$ = 18340.2 g

The actual yield of $C_2H_6$ = 14.0 kg = 14000 g (1 kg = 1000 g)

Now we have to calculate the percent yield of $C_2H_6$

$\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%$

Therefore, the percent yield of the reaction is, 76.34 %

5 0

4 years ago

Balance the following chemical reaction. ____NaOH+____H2SO4 → ___Na2SO4+____H2O

Balance the following chemical reaction. __NaOH+H2SO4 → _Na2SO4+____H2O