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aalyn [17]
3 years ago
5

If you have a sample of a solid with a mass of 48.2 g and a density of 14.3 g/cm3, what would the volume be? (round you answer t

o the nearest hundredth)
Chemistry
1 answer:
Alecsey [184]3 years ago
5 0

Answer:

mass is equal to density into volume

48.2=14.3*v

v=48.2/14.3

v=3.37

so volume is 3.40

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Yakvenalex [24]

Answer:

Explanation:

8.61+5.779 = 14.389 = 1.4389 × 10^1

25 - 12.5 = 1.25 x 10^1

56.35 / 13.2 = 4.2689

6 0
2 years ago
I got stood up by this guy<br> B R U H
ludmilkaskok [199]
What happened after that ? If you don’t mind me asking .
7 0
3 years ago
Read 2 more answers
Calculate the vapor pressure in torr of a solution containng 24.5 g of glycerin (C3H8O3) in 135 mL water at 30.0* C; the vapor p
Otrada [13]
Psolution = X · PH_20
= 0.966 · 31.8 torr
= 30.7 torr
3 0
2 years ago
By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask
tangare [24]

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = V_1=11.00 mL

Concentration of stock solution = M_1=0.823 M

Volume of stock solution after dilution = V_2=50.00 mL

Concentration of stock solution after dilution = M_2=?

M_1V_1=M_2V_2 ( dilution )

M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L

0.823 Molar is the molarity of the solution.

6 0
3 years ago
I have attached all the problems, but really if you just do one so I understand how to do it, that would be great!
Ivahew [28]

Answer:

Answer of question a is 345J.

Explanation:

In question a following is given in data:

-mass of iron (m) = 10.0 g

-temperature (ΔT) = final temperature- initial temperature= 100-25=  75 degree Celsius

-Specific Heat capacity of iron (c)= 0.46J/g°C.

Heat (Q)=?

Solution:

Formula for Heat is :

Q=m x c x ΔT

Q= 10 x 0.46 x 75

Q= 345 J.  

so, 345 joules of heat is needed to increase the temperature of 10 grams of iron.

  • From the above formula all other questions can easily be solved from the same procedure.
4 0
3 years ago
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