What is the molarity of a solution of 10% by mass cadmium sulfate, CdSO4 (molar mass = 208.46 g/mol) by mass? The density of the solution is 1.10 g/mL.

a. 0.528 M
b. 0.436 M
c. 0.479 M
d. 0.048 M
e. 22.9 M

Answers

Answer 1

Answer:

a. 0.528 M .

Explanation:

Hello!

In this case, since the given by-mass percent can be written as:

[tex]\frac{10gCdSO_4}{100g\ sol}[/tex]

By using the density and molar mass of the solute, cadmium sulfate, we can compute the molarity, by also making sure we convert from mL to L of solution:

[tex]M=\frac{10gCdSO_4}{100g\ sol}*\frac{1molCdSO_4}{208.46gCdSO_4} *\frac{1.10g\ sol}{1mL\ sol}*\frac{1000mL}{1L} \\\\ M=0.528M[/tex]

Thereby, the answer is a. 0.528 M .

Best regards.

Answer 2

The molarity of the solution of 10% by mass cadmium sulfate [tex](CdSO_4)[/tex] is approximately 0.479 M. The correct option is C.

To calculate molarity we need to find out how many moles of CdSO4 are present in the solution.

Given:

Mass of [tex]CdSO_4[/tex]= 10% by mass of the solutionMolar mass of [tex]CdSO_4[/tex] = 208.46 g/molDensity of the solution = 1.10 g/mL

We need to calculate the mass of [tex]CdSO_4[/tex]:

Mass of [tex]CdSO_4[/tex] = (10% / 100%) * Total mass of the solution

Mass of [tex]CdSO_4[/tex] = (10 / 100) * 1000 g (since the volume is 1 L, and the density is 1.10 g/mL)

Mass of [tex]CdSO_4[/tex] =  100 g

So, the number of moles of CdSO4:

Number of moles of [tex]CdSO_4[/tex] = Mass of CdSO4 / Molar mass of CdSO4

Number of moles of [tex]CdSO_4[/tex] = 100 g / 208.46 g/mol

Number of moles of [tex]CdSO_4[/tex] ≈ 0.479 moles

Then, we calculate the molarity of the solution:

Molarity = Number of moles of CdSO4 / Volume of the solution (in liters)

Molarity = 0.479 moles / 1 L

Molarity ≈ 0.479 M

Hence, the molarity of the solution of 10% by mass cadmium sulfate [tex](CdSO_4)[/tex] is approximately 0.479 M. The correct option is C.

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Explanation:

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Explanation:

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I hope this helps!

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Answers

0.042 moles of Hydrogen evolved

Further explanation

Given

I = 1.5 A

t = 1.5 hr = 5400 s

Required

Number of Hydrogen evolved

Solution

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Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻

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Answer:

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Answers

Answer:  Approximately [tex]0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}[/tex]

===================================================

Work Shown:

We have the following variables

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note: the symbol [tex]\Delta[/tex] is the uppercase Greek letter delta. It represents the difference or change in a value.

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[tex]Q = m*c*\Delta t\\\\714 = 52*c*51.5\\\\714 = 52*51.5*c\\\\714 = 2678*c\\\\2678*c = 714\\\\c = \frac{714}{2678}\\\\c \approx 0.26661687826737\\\\c \approx 0.267\\\\[/tex]

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Answer:

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Explanation:

Answer:

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Answers

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Answers

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Answers

Answer:

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Answer: C.

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Covalent bonding occurs between atoms when 1. The potential energy of the shared elec- trons is lower in the bond than it would be if the electrons were not shared between two nuclei 2. The potential energy of the shared elec- trons is independent of the covalent bond, but the potential energy of the newly formed molecule is greater than the individual atoms 3. Covalent bonding occurs when the octet rule is satisfied regardless of the potential en- ergy of the shared electrons 4. The potential energy of the shared elec- trons is higher in the bond than it would be if the electrons were not shared between two nuclei

Answers

Answer:The potential energy of the shared elec- trons is lower in the bond than it would be if the electrons were not shared between two nuclei

Explanation:

The entire essence of chemical combination is the attainment of a lower potential energy.

When two atoms are isolated at a large distance from each other, the potential energy is high.

However as the nuclei of the two atoms approach each other, the potential energy decreases until a minimum is attained. The distance between the nuclei at this point is the bond distance.

Hence, the potential energy of the shared electrons is lower in the bond than it would be if the electrons were not shared between two nuclei.

2.92 x 10^-15/
(5.6 x 10^-3) (4.16 x 10^9)

Answers

Answer:

3645.46

Explanation:

calculated

A student is trying to determine the density of a piece of metal. She uses a balance and finds that the mass of the piece of metal is 15.36 g. To determine the volume of the metal she fills a graduated cylinder to 30.0 mL and then drops the piece of metal into the graduated cylinder. The final volume of the metal 34.1 mL. What is the density of the piece of metal

Answers

Answer:

3.746 g/mL

Explanation:

Mass of metal =  15.36 g

Initial Volume = 30.0 mL

Final Volume = 34.1 mL

Volume of metal = Final Volume - Initial Volume = 34.1 mL - 30.0 mL

Volume of metal = 4.1 mL

Density = ?

The relationship between the values is given as;

Density = Mass of metal / Volume of metal

Inserting the values;

Density = 15.36 / 4.1  = 3.746 g/mL

The density of the piece of metal is 3.746 g/ml.

How to find the density of any metal?

Given the mass of metal is 15.36 g

The initial volume of a cylinder is 30.0 ml

The final volume after putting metal in it is 34.1ml

Step 1-  To find the volume:

The volume of metal = Final Volume – Initial Volume 30.0 mL

Volume of metal = 34.1 – 30.0 = 4.1 ml

Volume of metal = 4.1 mL

Step 2- Find the Density:

Density = Mass of metal is divided by the Volume of metal

[tex]\bold{Density = \dfrac{15.36 }{4.1} =3.746 g/ml}[/tex]  

Thus, the density is 3.746 g/ml.

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A patient is prescribed 100mg/day of antibiotic for 2 weeks. The antibiotic is available in vials that contain 20mg/vial of the drug. How many vials are necessary for the entire treatment?

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i have no idea, but i’m pretty sure you gotta divide
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