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3 years ago

What volume of a 1.0 M HCl is required to completely neutralize 25.0 ml of a 1.0 M KOH?

1 answer:

3 0

x 0,025
HCl+KOH--> KCl + H2O
1 1

c=1M
V=25ml=0,025l ( you need the volume to be in l)
c = n/V
n=c*V
n=1*0,025=0,025 m KOH

x=0,025m
c=1 M
V=n/c
V=0,025/1=0,025l=25ml HCl

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Please help me!!!!!!

Leona [35]

The answer is B

To write the equilibrium constant for an equation, all you have to do is divide the products by the reactants. The reactants are always on the left side, and the products are always on the right side. The coefficients of the elements will be written as the exponent of that same element. However, in this equation, we do not have to write any exponents, as there are no coefficient but 1.

5 0

3 years ago

If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4

statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

Step 1: Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

Step 2: Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

Step 3: Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Calculate Q

Q = [Ca2+] [SO42-]

[Ca2+]= 0.050 M [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

Step 7: Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

Step 8: Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

5 0

3 years ago

The nucleus’ pull on its electrons, taking into account any attractions weakened by shielding is called the:

Sever21 [200]

Answer:

It's called the shielding effect

Explanation:

It describes the decrease in attraction between an electron and the nucleus in any atom with more than one electron shell. (I hope this helps)

8 0

3 years ago

Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H

solmaris [256]

Answer:

36.37% is the percent yield of the reaction.

Explanation:

$4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)$

1)0.650 L nitrogen gas , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T = 295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol$

2) Moles of ammonia gas= $\frac{2.53 g}{17 g/mol}=0.1488 mol$

Moles of oxygen gas = $\frac{3.53 g}{32 g/mol}=0.1101 mol$

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

$\frac{4}{3}\times 0.1101 mol=0.1468 mol$ of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

$\frac{2}{3}\times 0.1101 mol=0.0734 mol$ of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

$\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the reaction:

$\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%$

36.37% is the percent yield of the reaction.

3 0

3 years ago

Determine the mass in grams of each nacl solution that contains 1.5g of nacl

Rama09 [41]

Answer:- 3333 g of solution.

Some of the question part is missing here. It would be like, "Determine the mass in grams of each NaCl solution that contains 1.5 g of NaCl.

(i) 0.045% NaCl by mass

Solution:- 0.045% NaCl by mass means 0.045 g of NaCl are present in 100 g of solution. 1.5 g of NaCl would be present in how many grams of solution?

We could solve this using proportions...

(0.045/100) = (1.5/X)

0.045(X) = 1.5(100)

0.045X = 150

X = 150/0.045 = 3333

So, 1.5 g of NaCl is present in 3333 g of solution.

7 0

4 years ago