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sashaice [31]
3 years ago
7

Below are 10 different metric prefixes and their abbreviations. Match them up. :)

Chemistry
1 answer:
Sladkaya [172]3 years ago
7 0

Answer:

1.G → giga  

2.da → deca

3.n → nano

4.m → milli

5.M → mega

6.k → kilo

7.c → centi

8.d → deci

9.h → hecto

10.µ → micro

Explanation:

In descending order:

1.G → giga (1 . 10⁹)  

5.M → mega (1 . 10⁶)

6.k → kilo (1 . 10³)

9.h → hecto (1 . 10²)

2.da → deca (1 . 10¹)

8.d → deci (1 . 10⁻¹)

7.c → centi (1 . 10⁻²)

4.m → milli (1 . 10⁻³)

10.µ → micro  (1 . 10⁻⁶)

3.n → nano (1 . 10⁻⁹)

In example:

1000000 m = 1 Mm

0,000000001 m = 1 nm

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A fast-moving neutron strikes a U235 nucleus. The nucleus shatters producing Sm155, Zn78, and three neutrons. Describe this chan
Natasha2012 [34]

The balanced nuclear equation for the reaction is

<h3>²³⁵₉₂U + ¹₀n —> ¹⁵⁵₆₂Sm +  ⁷⁸₃₀Zn + 3(¹₀n)</h3>

From the question given above, we were told that:

<u>A fast-moving neutron strikes a ²³⁵U nucleus. The nucleus shatters producing ¹⁵⁵Sm, ⁷⁸Zn, and three neutrons.</u>

The nuclear equation can be written as follow:

Neutron => ¹₀n

Uranium => ²³⁵₉₂U

Samarium => ¹⁵⁵₆₂Sm

Zinc => ⁷⁸₃₀Zn

Uranium + neutron —> Samarium + Zinc + 3 moles of neutron

<h3>²³⁵₉₂U + ¹₀n —> ¹⁵⁵₆₂Sm +  ⁷⁸₃₀Zn + 3(¹₀n)</h3>

 

 The nuclear equation above is balanced.

Learn more: brainly.com/question/9943790

3 0
3 years ago
I really need help on this it’s a major grade!!
Murrr4er [49]
Juan is taking out a loan for $2,500 with an annual compound interest rate of 7% for 3 years Juan will not make any additional deposits or withdrawals how much interest will he have to pay at the end of the loan period.
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The rapid decomposition of sodium azide, NaN3, to its elements is one of the reactions used to inflate airbags: 2 NaN3 (s)  2 N
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<u>Answer:</u> The amount of nitrogen gas produced is 3.864 grams

<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)

Given mass of NaN_3 = 6.00 g

Molar mass of NaN_3 = 65 g/mol

Putting values in equation 1, we get:

\text{Moles of }NaN_3=\frac{6.00g}{65g/mol}=0.092mol

For the given chemical reaction:

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

By Stoichiometry of the reaction:

2 moles of NaN_3 produces 3 moles of nitrogen gas

So, 0.092 moles of NaN_3 will produce = \frac{3}{2}\times 0.092=0.138mol of nitrogen gas

Now, calculating the mass of nitrogen gas by using equation 1, we get:

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.138 moles

Putting values in equation 1, we get:

0.138mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.138mol\times 28g/mol)=3.864g

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3 years ago
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Answer:

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