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lawyer [7]
3 years ago
6

What is 7.09x10 to the 5 minus 6.10xto the 4 in scientific notation

Chemistry
1 answer:
STALIN [3.7K]3 years ago
4 0
When subtracting scientific notation, you first need to make the exponents the same. The standard is to make the smaller exponent the same as the higher one.
Let's take a look at your given:

7.09 x 10^{5}
6.10 x 10^{4} This is smaller. 

Every time you add to the exponent, you need to move the decimal. So to make it the same, you need to add 1 to the 4 to get 5. Then move the decimal to the left. Your new notation would be:
.610 x 10^{5}

Then you just line up the decimal as you would before subtracting and just copy the base and exponent.
7.09 x 10^{5}
- .61 x 10^{5}
=
6.48 x 10^{5}
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A snowflake contains 2.1 10^18 molecules of water. How many moles of water does it contain?
IgorC [24]

Answer:

divide the # of molecules by avogadros number and get 3.48 x 10^-6

Explanation:

6 0
2 years ago
A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt
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Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

<u>Step 1:</u> Data given

Sample 1: The 1.15 M sample  has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1:  1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

<u>Step 3:</u> Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams  ( in 55mL)

<u>Step 4</u>: Calculate moles of AgBr

AgNO3 reacts with KBr  

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0
3 years ago
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