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2 years ago

What is the concentration of bromide, in ppm, if 115.91 g MgBr2 is dissolved in 1.31 L water.

Chemistry

1 answer:

Ksenya-84 [330]2 years ago

6 0

Answer:- Bromide concentration is 80748 ppm.

Solution:- ppm is milligram per liter. Grams of magnesium bromide are given and we are asked to calculate the ppm concentration of bromide ion, So, let's first calculate the grams of bromide and then divide it by the given volume.

Molar mass of Magnesium bromide = $24.305+2(126.90)=278.105$

$115.91gMgBr_2(\frac{1molMgBr_2}{278.105gMgBr_2})(\frac{2molBr}{1molMgBr_2})(\frac{126.90gBr}{1molBr})$

= 105.78 g Br

Let's convert grams of Br to milligrams:

since, 1 g = 1000 mg

$105.78g(\frac{1000mg}{1g})$

= 105780 mg

concentration of Br = $\frac{105780mg}{1.31L}$

= 80748 mg per liter

Since, 1 ppm = 1 mg per liter

So, the bromide concentration is 80748 ppm.

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A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and

Ilya [14]

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

$Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-$

Molarity of the silver nitrate solution = 0.03337 M

Volume of the silver nitrate = 23.28 mL = 0.02328 L

Moles of silver nitrate = n

$0.03337 M=\frac{n}{0.02328 L}$

n = $0.03337 M\times 0.02328 L=0.0007768 mol$

According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.

Then 0.0007768 moles of silver nitrate will react with:

$\frac{1}{1}\times 0.0007768 mol=0.0007768 mol$ chloride ions.

In one mole of aldrin there are 6 moles of chloride ions.

Then moles of aldrin containing 0.0007768 moles chloride ions are:

$\frac{0.0007768 mol}{6}=0.0001295 mol$

Moles of aldrin present in the sample = 0.0001295 mol

Mass of 0.0001295 moles of aldrin present in the sample :

0.0001295 mol × 364.92 g/mol =0.04726 g

Percentage of an aldrin in the sample:

$\frac{0.04726 g}{0.1064 g}\times 100=44.41\%$

8 0

2 years ago

Give the product for the reaction of 1-butene with methanol in the presence of acid. The mechanism is the same as the mechanism

Nuetrik [128]

Answer:

The final result is 2-methoxybutane.

Explanation:

1-butene has a carbon-carbon double bound between C1 and C2.

When it will react with methanol, in the presence of an acid, the result will be an ether.

C4H8 + CH3OH → C5H12O

An acid-catalyzed ether synthesis from alkenes is limited by carbocation stability.

In the first step, the double bound will disappear. The C2 atom will be a C+ atom, this becaus it has only 3 bounds and not 4.

This C+ -atom will atract the O- atom to form an ether. The CH3 of methanol will bind on the C3 atom, this is the most stable position.

2-methoxybutane will be formed. It has a structural formula of C5H12O

1-methoxybutane will not be formed, because it's less stable.

1-butanol will be formed when water is added to 1-butene. The mechanism has the same principle but not the same product.

1-ethoxybutane and 2-ethoxybutane have a structural formula of C6H14O, this will not be the final result.

The final result is 2-methoxybutane.

8 0

2 years ago

A 20.0 -\,L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.2 atm. Enough weight is suddenly removed fr

zzz [600]

Answer:

1. ΔE = 0 J

2. ΔH = 0 J

3. q = 3.2 × 10³ J

4. w = -3.2 × 10³ J

Explanation:

The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.

The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.

P₁ × V₁ = P₂ × V₂

3.2 atm × 20.0 L = 1.6 atm × V₂

V₂ = 40 L

The work (w) can be calculated using the following expression.

w = - P . ΔV

where,

P is the external pressure for which the process happened

ΔV is the change in the volume

w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J

The change in the internal energy is:

ΔE = q + w

0 = q + w

q = - w = 3.2 × 10³ J

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