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Bumek [7]
3 years ago
11

PLEASE HELP ME!!!! ILL FIND A WAY TO GET YOU 100 POINTS!!!

Chemistry
1 answer:
Umnica [9.8K]3 years ago
8 0

It’s c had this problem last week

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Determine the mass of copper that has the same number of atoms as there are in 6.95 mg of potassium. Answer in units of mg.
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3 years ago
What volume will be occupied by 33.0 grams of CO2 at 500 torr and 27 °C?
wlad13 [49]

Answer:

V = 27.98 L

Explanation:

Given data:

Mass of CO₂ = 33.0 g

Pressure = 500 torr

Temperature = 27°C

Volume occupied = ?

Solution:

Number of moles of CO₂:

Number of moles = mass/molar mass

Number of moles = 33.0 g/ 44 g/mol

Number of moles = 0.75 mol

Volume of CO₂:

PV = nRT

R = general gas constant = 0.0821 atm.L/ mol.K  

Now we will convert the temperature.

27+273 = 300 K

Pressure = 500 /760 = 0.66 atm

By putting values,

0.66 atm×V = 0.75 mol × 0.0821 atm.L/ mol.K  × 300 K

V = 18.47 atm.L/0.66 atm

V = 27.98 L

4 0
2 years ago
46g of a mineral contained 16g copper, 14g iron and 16g sulphur. Calculate the empirical
pochemuha

Answer:

CuFeS2

Explanation:

Calculate the moles of each substance by doing moles= mass/relative atomic mass. you should get 0.25 moles of copper and iron and 0.5 moles of sulfur. Then divide all of those numbers by 0.25 (as its the lowest value) you should get 1 for copper and iron and 2 for sulfur. This represents the ratio that they are in within the mineral.

7 0
3 years ago
A solenoid is a coil of wire that produces a magnetic field when carrying an electric current. The strength of the magnetic fiel
BaLLatris [955]

Answer: Motion, providing a burst of power that can move a specific part of the device.

Explanation: Hope this helps

3 0
2 years ago
Read 2 more answers
A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b
sveta [45]

Answer:

  • Part A: m = 0.02356 mol/kg = 0.02356 m
  • Part B: Xsolute = 4.243×10⁻⁴
  • Part C: % m/m = 0.1376%
  • Part D: ppm = 1,376 ppm

Explanation:

<u>1. Data:</u>

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

<u>2. Formulae</u>

  • M = n solute / V sol (L)
  • m = n solute / Kg solvent
  • X solute = n solute / N total
  • % m/m = (mass of solute / mass of solution) × 100
  • ppm = (mass of solute / mass of solution) × 1,000,000
  • density = mass in grams / volume in mL

<u>3. Solution</u>

<u>Part A: Calculate the molality of the salt solution. </u>

<u />

            m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg  H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

<u>Part B: Calculate the mole fraction of salt in this solution</u>.

          X solute = n solute / N total

i) n solute =  2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

<u>Part C: Calculate the concentration of the salt solution in percent by mass</u>.

         % m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

       ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

5 0
3 years ago
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