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Usimov [2.4K]
3 years ago
7

Students working in lab use 52.3g of CaCl2 to make a solution. When the solution is brought to a final volume of 800. mL, what i

s the concentration?
A. 0.139 M
B. 0.153 M
C. 0.588 M
D. 0.632 M
Chemistry
1 answer:
PolarNik [594]3 years ago
6 0

Answer:

The answer is C: .588 M

Explanation:

[CaCl2] = mols CaCl2/Liters of solution

molar mass of CaCl2 = 110.98 g/ 1 mole   (from periodic table)

800 mL/1000 mL = .8 Liters of solution

1)

52.3g CaCl2/ 110.98g = .47 mole

2)

.47 mole/.8 L = .589 M

So the answer is C

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Explanation:

Building Vocabulary

Match each term with its definition by writing the letter of the correct definition on

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5. nucleus   b

6. proton     f

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8. electron  d

9. atomic number    g

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11. mass number      a

12. energy level       e

a. the sum of protons and neutrons in the nucleus of an

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e. a specific amount of energy related to the movement

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2 years ago
Cuando un gas que se encuentra a 20°c se calienta hasta los 40°c sin que varie su presion, su volumen se duplica
kati45 [8]

Answer:

Not doubled

Explanation:

The equation below represent the ideal gases relationship

PV ÷ T = constant

Here

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T denotes temperature in degrees Kelvin

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In the case when the temperature would be determined in degrees celsius at 0 degrees so the volume would be zero

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Do the number of moles of a gas impact the pressure of a gas?
lapo4ka [179]
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3 years ago
Read 2 more answers
When active metals such as magnesium are immersed in acid solution, hydrogen gas is evolved. Calculate the volume of H2(g) at 30
V125BC [204]

Answer:

The volume of  H₂ (g) obtained is 22.4L

Explanation:

First of all, think the reaction:

2HCl (aq) + Zn (s) → ZnCl₂ (aq)  + H₂ (g)

You have to add a 2, in the HCl to get ballanced.

Now we should know how many moles of each reactant, do we have.

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Notice that volume is in mL, so I must convert to L.

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My limiting reactant is the HCl, for 0.764 moles of Zinc, I need 1.528 (0.764 .2) of HCl, and I only have 0.2 moles.

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P. V = n . R . T

2 atm . V = 0.2 mol . 0.08206L atm/K mol . 273K

V =  (0.2 mol . 0.08206L atm/K mol . 273K ) / 2 atm

V = 2.24 L

4 0
3 years ago
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