All categories

3 years ago

What is the solution to the problem expressed to the correct number of significant figures?

Chemistry

1 answer:

Fed [463]3 years ago

8 0

Answer:

(102 900 ÷ 12) + (170 × 1.27) = 8800

Step 1. Evaluate the expressions inside the parentheses (PEMDAS)

102 900 ÷ 12 = 8575

170 × 1.27 = 215.9

In multiplication and division problems, your answer can have no more significant figures than the number with the fewest significant figures.

Thus, the underlined digits are not significant, but we keep them in our calculator to avoid roundoff error.

Step 2. Do the addition (PEMDAS).

8575

+ 215.9

= 8790.9

Everything that you add to an insignificant digit gives an insignificant digit as an answer.

Thus, the underlined digits are not significant.

We must drop them and round up the answer to 8800.

Explanation:

You might be interested in

What is Pb+FeSO4-->PbSO4+Fe balanced and the type of reaction?

Evgesh-ka [11]

Answer:

It's simple substitution reaction and there is no balance for the equation. It's right written.

Explanation:

8 0

3 years ago

Why is it important to be careful when walking on an icy sidewalk?

Archy [21]

Answer:

The is less friction on an icy surface

Explanation:

This is because when u walk on plain ground there is steady friction on the surface but when u walk on an icy floor these is barely any friction which is why it's hard to walk on.

4 0

2 years ago

Read 2 more answers

What is a characteristic of nuclear fusion

coldgirl [10]

75% hydrogen
25% helium
2%carbon, nitrogen

8 0

3 years ago

Although coal is a complex mixture of substances, its elemental composition can be approximated by the formula <img src="https:/

Svetllana [295]

Answer:

The correct answer is option b.

Explanation:

$2C_{135}H_{96}O_9NS+313O_2\rightarrow 270CO_2+2NO_2+2SO_2+96H_2O$

Mass of coal burned = $1.00\times 10^6$ metric ton = 1.00\times 10^6\times 10^3 kg=10^9 kg[/tex]

1 metric ton = $10^3kg$

Molar mass of coal :

$135\times 12g/mol+96\times 1g/mol+9\times 16 g/mol+1\times 14 g/mol+1\times 32 g/mol=1906 g/mol=1.906 kg/mol$

1 g = 0.001 kg

Moles of coal ,n : $\frac{10^9 kg}{1.906 kg/mol}=5.247\times 10^8 mol$

If 2 moles of coal on combustion gives 270 moles of carbon dioxide than n moles of coal will give;

$5.247\times 10^8 mol\times \frac{270}{2}=7.083\times 10^{10} mol$ of carbon dioxide.

Molar mass of carbon dioxide gas = 44 g/mol = 0.044 kg/mol

Mass of $7.083\times 10^{10} mol$ of carbon dioxide:

$7.083\times 10^{10} mol\times 0.044 kg/mol=3.11\times 10^{11} kg$

5 0

4 years ago

How many moles of k3po4 can be formed when 4.4 moles of h3po4 react with 3.8 moles of koh? h3po4 + koh yields h2o + k3po4 be sur

schepotkina [342]

the balanced chemical equation for the reaction is as follows

H₃PO₄ + 3KOH ---> K₃PO₄ + 3H₂O

stoichiometry of H₃PO₄ to KOH is 1:3

first we have to find which the limiiting reactant is

as the amount of product formed depends on the amount of limiting reactant present

number of H₃PO₄ moles reacted - 4.4 mol

if H₃PO₄ is the limiting reactant

1 mol of H₃PO₄ reacts with 3 mol of KOH

then 4.4 mol of H₃PO₄ reacts with - 3 x 4.4 mol = 13.2 mol of KOH

but only 3.8 mol of KOH is present

therefore KOH is the limiting reactant

stoichiometry of KOH to K₃PO₄ is 3:1

number of KOH moles reacted - 3.8 mol

therefore number of K₃PO₄ formed = number of KOH moles reacted / 3

= 3.8 mol / 3 = 1.3 mol

answer is 1.3 mol of K₃PO₄

3 0

4 years ago