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3 years ago

How many mL of a 2.0M NaBr solution are needed to make 200.0 mL of 0.50M NaBr?

2 answers:

8 0

This is a classical C1V1 = C2V2 question. We are given the initial concentration (2.0M), the final concentration (0.5M) and the final volume (200mL). We are asked how many mL of 2.0 M NaBr we require to make up the final concentration and volume, so we are calculating the initial volume (V1)

Therefore:
C1V1 = C2V2
2.0M(V1) = (0.5M)(200.0 mL)
V1 = 50 mL
So 50 mL of NaBr is needed to make a dilution to 0.5M in 200 mL.

ololo11 [35]3 years ago

6 0

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

2.0 M x V1 = 0.50 M x 200 mL

V1 = 50 mL of 2.0M NaBr is needed

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The decay of the isotope iodine-131 is first-order with a rate constant of 0.138 d−1. All radioactive decay is first order. How

Allisa [31]

Answer:

16.7 days

Explanation:

We are given;

A radioactive isotope Iodine-131`

The decay rate is 0.138 d⁻¹

The percent decayed is 90%

We are suppose to calculate the number of days for the decay.

Using the formula;

$In(\frac{[A_{0}]}{[A]})=kt$

Where, $[A_{0}]$ is the initial concentration and $[A]$ is the new concentration.

Rearranging the formula;

$t=In(\frac{[A_{0}]}{[A]})/k$

Assuming the initial concentration is x, then the final concentration after 90% decay will be 0.10x

Therefore;

$t=In(\frac{[x]}{[0.10x]})(\frac{1}{0.138})$

$t=In(\frac{[1]}{[0.10]})(\frac{1}{0.138})$

$t=In(10.0)(\frac{1}{0.138})$

$t=(2.3026)(\frac{1}{0.138})$

$t=16.685$

Time = 16.7 days

Therefore, it will take 16.7 days for 90% of I-131 to decay to Xe-131

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4 years ago

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Alina [70]

Answer:

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Explanation:

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3 years ago

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What oxidation state(s) can the alkaline earth metals exhibit?

iragen [17]

Answer:

Explanation:

Alkaline earth metals are present in group 2 of periodic table. There are six elements in second group. Beryllium, magnesium, calcium, strontium, barium and radium.

All have two valance electrons.

Electronic configuration of Beryllium:

Be = [He] 2s²

Electronic configuration of magnesium.

Mg = [Ne] 3s²

Electronic configuration of calcium.

Ca = [Ar] 4s²

Electronic configuration of strontium.

Sr = [Kr] 5s²

Electronic configuration of barium.

Ba = [Xe] 6s²

Electronic configuration of radium.

Ra = [ Rn] 7s²

They are present in group two and have same number of valance electrons (two valance electrons) and show oxidation state +2 by loosing two valance electrons. They also show similar reactivity.

They react with oxygen and form oxide.

2Ba + O₂ → 2BaO

2Mg + O₂ → 2MgO

2Ca + O₂ → 2CaO

this oxide form hydroxide when react with water,

BaO + H₂O → Ba(OH)₂

MgO + H₂O → Mg(OH)₂

CaO + H₂O → Ca(OH)₂

With sulfur,

Mg + S → MgS

Ca + S → CaS

Ba + S → BaS

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3 years ago

A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spheric

Luba_88 [7]

Complete Question

A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spherical air tank that measures 74.0 wide. The biologist estimates she will need 2600 L of air for the dive. Calculate the pressure to which this volume of air must be compressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to significant digits.

Answer:

The pressure required is $P_2= 12.2 \ atm$