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3 years ago

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Chemistry

1 answer:

kykrilka [37]3 years ago

7 0

Answer:b yes b yes c no c yes d yes d no

Explanation: wanna trade r o b l o x acc

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What is the percent yield of the purification system

Jlenok [28]

Answer: 90.3

Explanation:

3 0

3 years ago

A student needs to prepare 100. mL of 0.612 M Cu(NO3)2 solution. What mass, in grams, of copper(II) nitrate should the student u

Temka [501]

Answer: 11.5 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution

$Molarity=\frac{n\times 1000}{V_s}$

where,

Morality = 0.612 M

n= moles of solute

$V_s$ = volume of solution in ml = 100 ml

Now put all the given values in the formula of molarity, we get

$0.612=\frac{n\times 1000}{100ml}$

$n=0.0612moles$

$Mass={\text {moles of solute }}{\times {\text {molar mass}}=0.0612moles\times 187.56g/mol=11.5g$

Therefore, the mass of copper (II)nitrate required is 11.5 grams

3 0

3 years ago

200.0 mL of 3.85 M HCl is added to 100.0 mL of 4.6 M barium hydroxide. The reaction goes to completion. What is the concentratio

Ede4ka [16]

Answer:

2.387 mol/L

Explanation:

The reaction that takes place is:

2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

First we calculate how many moles of each reagent were added:

HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl

Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂

460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.

Now we calculate how many moles of Ba(OH)₂ reacted, by converting the total number of HCl moles to Ba(OH)₂ moles:

203.85 mmol HCl * $\frac{1mmolBa(OH)_{2}}{2mmolHCl}$ = 101.925 mmol Ba(OH)₂

This means the remaining Ba(OH)₂ is:

460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂

There are two OH⁻ moles per Ba(OH)₂ mol:

OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻

Finally we divide the number of OH⁻ moles by the total volume (100 mL + 200 mL):

716.15 mmol OH⁻ / 300.0 mL = 2.387 M

So the answer is 2.387 mol/L

7 0

3 years ago

A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a

KatRina [158]

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

Partial pressure of methane (pCH₄): 431 mmHg

Partial pressure of carbon dioxide (pCO₂): 504 mmHg

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461

XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

3 0

3 years ago

Oxana [17]

Answer:

1422mg of acetaminophen

Explanation:

The maximum dose of acetaminophen is 15.0 mg of acetaminophen per kg of person.

To know the maximum single dosage of the person we need to convert the 209lb to kg (Using 1kg = 2.2046lb):

209lb * (1kg / 2.2046lb) = 94.8

The person weighs 94.8kg and the maximum single dosage for the person is:

94.8kg * (15.0mg acetaminophen / kg) =

1422mg of acetaminophen

4 0

3 years ago

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