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Georgia [21]
3 years ago
6

How do you write 145,000,000 in scientific notation?

Chemistry
1 answer:
belka [17]3 years ago
5 0

Answer:

\boxed{1.45 \times 10^{8}}

Explanation:

A number in scientific notation has the form N × 10ⁿ

where N is a decimal number called the mantissa and n is an integer called the exponent.

We must have 1 ≤ N < 10

Step 1. Determine the mantissa

Move the decimal place to the left to create a new number between 1 and 10.

145 000 000 ⟶ 1.450 000 00; N = 1.45

Step 2. Determine the exponent.

The exponent is the number of times you moved the decimal to get the mantissa.

You moved the decimal eight places to the left, so the exponent n = 8.

Step 3. Write the number in scientific notation

The number in scientific notation is \boxed{1.45 \times 10^{8}}.

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3 years ago
Water is a polar molecule, meaning the electrons that are shared between the atoms aren't shared equally so oxygen attracts more
amm1812

Answer:

Magnet with a positive and a negative pole

Explanation:

A great analogy to demonstrate what a polar molecule looks like is to imagine a magnet. A magnet has one positively charged end and one negatively charged end, two poles, that is.

Imagine that we have a magnet of a shape of a prism (water molecule has a bent shape). The two base vertices of the face of the triangle are positively charged, that's because hydrogen is less electronegative than oxygen and, hence, the two hydrogen atoms are partially positively charged in a water molecule.

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3 years ago
In the following reaction, how many grams of oxygen will react with 10.47 grams of benzene (C6H6)? 2C6H6 + 1502 12CO2 + 6H2O The
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3 years ago
Which missing item would complete this beta decay reaction?
madam [21]

Answer:

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3 years ago
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
3 years ago
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