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choli [55]
3 years ago
6

What does it mean when your teacher wants you to round 143 to two significant figures

Chemistry
1 answer:
zhenek [66]3 years ago
5 0

Answer:The answer is 3...I think

Explanation:

143

Sig Figs

3

Decimals

0

Scientific Notation

1.43 × 102

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How many atoms does one formula unit of Sb4(SiO4)3 have
kotegsom [21]

Answer:

There are 19 atom sin total in antimony silicate, Sb_4(SiO_4)_3.

Explanation:

Number of antimony atoms = 4 × 1 = 4

Number of silicon atoms = 3 × 1 = 3

Number of oxygen atoms = 3 × 4 = 12

Number of atoms in antimony silicate:

= 4 + 3 + 12 = 19

There are 19 atom sin total in antimony silicate, Sb_4(SiO_4)_3.

6 0
3 years ago
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jasenka [17]

Answer:

Mitochondria - energy production in plant and animal tissue

Explanation:

The mitochondria is the power house of a cell. it contains adenosine-tri- phosphate or ATP molecules that produce energy through active transport. The mitochondria is responsible for energy production in both plant and animal cells

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3 years ago
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of N_2 = 100 kg  = 100000 g

Mass of H_2 = 100 kg = 100000 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 3571.43 moles of N_2 react with 3571.43\times 3=10714.29 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 3571.43 moles of N_2 react to give 3571.43\times 2=7142.86 moles of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0
3 years ago
A 30-kg girl and a 25-kg boy face each other on friction-free roller blades. the girl pushes the boy, who moves away at a speed
dsp73
0.83 m/s seems the correct answer, hope it helps
6 0
3 years ago
Determine the mass of 2.75 moles of CaSO4. Record your work and your answer.
maw [93]

Explanation:

RFM \:  = 160 \: g \\ 1 \: mole \: weighs \: 160 \: g \\ 2.75 \: moles \: weighs \:  \frac{2.75 \times 160}{1} g \\  = 440 \: g

6 0
3 years ago
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