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balandron [24]
3 years ago
5

a chemist encounters an unknown metal. They drop the metal into a graduated cylinder containing water, and find the volume chang

e is 3.2 mL. If the metal weighs 1.5g, what is the density in g/mL of the metal?
Chemistry
1 answer:
dybincka [34]3 years ago
6 0

Answer:

we know that the density is the mass of the object divided by its volume: d=m/v

Explanation:

1.5/3.3 = 0.4545454545 g/ml

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How many moles of water contain 1.3 x 10^24 molecules?
Vika [28.1K]
C is the correct answer
3 0
3 years ago
How many moles are equal to 1.3 x 1024 atoms of aluminum?
marshall27 [118]

Answer:

<h3>The answer is 2.16 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L}  \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{1.3 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\  = 2.159468...

We have the final answer as

<h3>2.16 moles</h3>

Hope this helps you

8 0
3 years ago
How did corana virus start<br><br><br>​
Veseljchak [2.6K]

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7 0
2 years ago
Read 2 more answers
When 250 ml of water is added to 35 ml of 0.2 M HCl
Paladinen [302]
Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.

You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume

C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL

C1 x 250 mL = 0.2 M x 35 mL

C1 = (0.2 M x 35 mL) / 250 mL

C1 = 0.028 M of water added to 35mL of 0.2M HCl

Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl
8 0
3 years ago
Read 2 more answers
Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

                                                                                           = 75.3%

4 0
3 years ago
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