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MariettaO [177]
3 years ago
9

Mass of 60.009 g and radius of 1.02 cm what is the density of the sphere in (g/ml)

Chemistry
1 answer:
alexandr402 [8]3 years ago
4 0
Density is given as mass / volume.

Mass is the sphere is 100 g.

Volume of the sphere = (pi∗r3)∗4/3
(
p
i
∗
r
3
)
∗
4
/
3

=(4∗22∗3∗3∗3)/(7∗3)cm3
=
(
4
∗
22
∗
3
∗
3
∗
3
)
/
(
7
∗
3
)
c
m
3

=792/7
=
792
/
7
cm3
3

Therefore, Density is 100/(792/7)g/cm3
100
/
(
792
/
7
)
g
/
c
m
3

Which gives: density = 0.883838 g/cm3
g
/
c
m
3

If you want to change the units to kg per cubic metres, then we need to divide this value by 1000( for g to kg) and multiply by 100 * 100 * 100 (for cm to m).

This makes the density to be 883.83 kg/m3
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What was the plum pudding atomic model
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2 years ago
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For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaOH, calculate:
Leviafan [203]
1) Chemical reaction

HCl        +       NaOH      --->      NaCl + H2O

25.0 ml            
0.150 M            0.250M

2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution

0.001875 mol HCl => 0.001875 mol H(+)

Volume = Volume of HCl solution + Volumen of NaOH solution added

Volume of HCl solution = 0.0250 l

Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l

Total volume = 0.0250 l + 0.0075 l = 0.0325 l

[H+] = 0.001875 mol / 0.0325 l = 0.05769 M

pH = - log [H+] = - log (0.05769) = 1.23

Answer: 1.23

3) Equivalence point

0.02500 l * 0.150 M = 0.250M * V

=> V = 0.02500 * 0.150 / 0.250 = 0.015 l

4) 1.00 ml NaOH added beyond the equivalence point

1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess

0.00025 mol NaOH = 0.00025 mol OH-

Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l

[OH-] = 0.00025 mol / 0.041 l = 0.00610 M

pOH = - log (0.00610) = 2.21

pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76

Answer: 11.76
6 0
3 years ago
What volume will 0.405 g of krypton gas occupy at STP?
Rufina [12.5K]

Answer:

The answer to your question is V = 0.108 L or 108 ml

Explanation:

Data

Volume = ?

mass = 0.405 g

Temperature = 273°K

Pressure = 1 atm

Process

1.- Convert mass of Kr to moles

                  83.8 g of Kr -------------------- 1 mol

                     0.405 g     -------------------  x

                     x = (0.405 x 1) / 83.8

                     x = 0.0048 moles

2.- Use the Ideal gas law to solve this problem

                   PV = nRT

- Solve for V

                      V = nRT / P

- Substitution

                      V = (0.0048)(0.082)(273) / 1

- Simplification

                       V = 0.108 / 1

- Result

                       V = 0.108 L

8 0
3 years ago
Read 2 more answers
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