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Over [174]
3 years ago
5

Perform the following operation

Chemistry
1 answer:
Charra [1.4K]3 years ago
3 0

Answer:

-2.7 ×10²

Explanation:

Scientific notation is the way to express the large value in short form.

The number in scientific notation have two parts.

. The digits (decimal point will place after first digit)

× 10 ( the power which put the decimal point where it should be)

for example the number 6324.4 in scientific notation will be written as = 6.3244 × 10³

Given measurements = 4.3×10² - 7.0 ×10²

= -2.7 ×10²

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nasty-shy [4]

Answer:

Ionic bond

Explanation:

Covalent bonds can only form between non-metals and copper is a metal so it would be an ionic bond.

6 0
3 years ago
3.79 x 10^8<br> =<br> .mol<br> 2<br> 2<br> kg.m<br> g.cm
yuradex [85]

Answer:

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3 0
3 years ago
If 6.00 g of the unknown compound contained 0.200 mol of C and 0.400 mol of H, how many moles of oxygen, O, were in the sample?
Arada [10]

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

 

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

 

Answer:

<span>0.2 moles O</span>

8 0
4 years ago
Read 2 more answers
When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio
nignag [31]

Answer:

3.73 mL

Explanation:

To solve this problem we can use the equation C₁V₁=C₂V₂, where:

  • C₁ = 6.57 M
  • V₁ = ?
  • C₂ = 0.0490 M
  • V₂ = 500 mL

We<u> input the data</u>:

  • 6.57 M * V₁ = 0.0490 M * 500 mL

And <u><em>solve for V₁</em></u>:

  • V₁ = 3.73 mL

So 3.73 mL of a 6.57 M stock solution would be required to prepare 500 mL of a 0.0490 M solution.

4 0
3 years ago
A gas is heated from 213.0 K to 298.0 K and the volume is increased from 14.0 liters to 35.0 liters by moving a large piston wit
konstantin123 [22]

Answer:

The answer is ""

Explanation:  

Given:

V_1= 14.0\ L\\\\P_1= 3.15\ atm\\\\T_1 = 213.0 \ K\\\\T_2= 298.0\ K\\\\V_2= 35.0\ L\\\\P_2 = ?

Using formula:

\to \frac{P_1V_2}{T_1} = \frac{P_2V_2}{T_2}\\\\P_2 = \frac{P_1V_1 T_2}{T_1 V_2}  \\\\

     = \frac{3.15 \times 14.0 \times  298.0}{213.0 \times 35.0}  \\\\= \frac{13141.0}{7455} \\\\=1.7627\\\\=1.8\ atm

5 0
3 years ago
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