4.What volume of 0.120 M HNO3(aq) is needed to
completely neutralize 150.0 milliliters of 0.100 M
NaOH(aq)?
A. 62.5 mL
B. 125 ml
C.
180. mL
D. 360. mL

Answers

Answer 1
Answer:

Answer:

B) 125 mL

Explanation:

M1V1=M2V2

(0.120M)(x)=(150.0 mL)(0.100M)

x= 125 mL

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Which substance yields hydroxide ion as the only negative ion in aqueous solution?

Answers

Missing question:
a. Mg(OH)2.
b. CH3Cl.
c. MgCl2 .
d. C2H4(OH)2.
Answer is: a. Mg(OH)₂.
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In this example magnesium hydroxide is an Arrhenius base:
Mg(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
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All light travels at the same speed.
O True
O False

Answers

False

thank me later

Answer:

False

Explanation:

It depends on wixh kid of light is it

Example

Sun light difders from lamp light

A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. What is the final molarity of the dilute solution?

Answers

A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. the final molarity of the dilute solution is 0.102 M.

From the question given above, the following data were obtained:

Volume of stock solution (V1) = 5 mL

Molarity of stock solution (M₁) = 5.103 M

Volume of diluted solution (V₂) = 250 mL

Molarity of diluted solution (M₂) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

5.103 × 5 = M2 × 250

25.515 = M2 × 250

Divide both side by 250

M2 = 25.515 / 250

M2 = 0.102 M

Thus, the molarity of the diluted solution is 0.102 M.

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Answer:

0.102 M.

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V1) = 5 mL

Molarity of stock solution (M1) = 5.103 M

Volume of diluted solution (V2) = 250 mL

Molarity of diluted solution (V2) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M1V1 = M2V2

5.103 × 5 = M2 × 250

25.515 = M2 × 250

Divide both side by 250

M2 = 25.515 / 250

M2 = 0.102 M

Thus, the molarity of the diluted solution is 0.102 M.

Why is it important to know what temperature scale is being used in a given situation?

Answers

It depends, for example, it is quite important to know the Kelvin scale (i.e 0 degrees Celsius is 273 K and -273 degrees Celsius is 0 K ) when dealing gases. But I don't know other situations where you would need to know other temperature scales.

Hope this helps and also if you are using Fahrenheit 1 Fahrenheit is -17.22 degrees Celsius 

Describe the environmental impacts that are involved during the design, research and development, and marketing phases of cell phones.

Answers

Answer:

joe mama

Explanation:

400 + 20 + 420

Suppose of lead(II) acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the lead(II) acetate is dissolved in it. Round your answer to significant digits.

Answers

Answer:

0.294 M

Explanation:

The computation of the final molarity of acetate anion is shown below:-

Lead acetate = Pb(OAc)2

Lead acetate involves two acetate ion.

14.3 gm lead acetate = Mass ÷ Molar mass

= 14.3 g ÷ 325.29 g/mol

= 0.044 mole

Volume of solution = 300 ml.

then

Molarity of lead is

= 0.044 × 1,000 ÷ 300

= 0.147 M

Therefore the molarity of acetate anion is

= 2 × 0.147

= 0.294 M

Final answer:

To calculate the final molarity of acetate anion in the solution, consider the dissociation of lead(II) acetate and the presence of ammonium sulfate. When ammonium sulfate is added, it reacts with the lead(II) cations, leaving only the acetate anions in the solution. The final concentration of acetate anions is therefore the same as the initial concentration.

Explanation:

To calculate the final molarity of acetate anion in the solution, we need to consider the dissociation of lead(II) acetate and the presence of ammonium sulfate. Lead(II) acetate will dissociate into lead(II) cations (Pb2+) and acetate anions (CH3COO-) in solution. However, when ammonium sulfate is added, the sulfate anions (SO42-) react with the lead(II) cations, forming lead(II) sulfate and removing them from solution. This leaves us with only the acetate anions.

First, calculate the concentration of the acetate anions in the lead(II) acetate solution. Then subtract the concentration of the acetate anions that reacted with the lead(II) cations to form lead(II) sulfate. This will give us the final concentration of acetate anions in the solution.

Let's assume we have an initial concentration of lead(II) acetate of X M. The dissociation of lead(II) acetate can be represented as:

Pb(CH3COO)2(s) ⇌ Pb2+(aq) + 2CH3COO-(aq)

Since we assume the volume of the solution doesn't change when the lead(II) acetate is dissolved, the initial concentration of acetate anions is also X M.

When ammonium sulfate is added, it reacts with the lead(II) cations according to the reaction:

Pb2+(aq) + SO4^2-(aq) ⇌ PbSO4(s)

Since the concentration of lead(II) sulfate is negligible, we can assume that all the lead(II) cations react with the sulfate anions. This removes the lead(II) cations from solution, leaving us with only the acetate anions.

Therefore, the final concentration of acetate anions is still X M.

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