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2 years ago

How many significant figures should the following answer have? (23.34 cm)(90076 cm)(2.1223 cm) = cm

Chemistry

2 answers:

bezimeni [28]2 years ago

8 0

Answer : The number of significant figures are, four (4).

Explanation :

The rule apply for the multiplication and division is :

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

As we are given :

$(23.34cm)\times (90076cm)\times (2.1223cm)$

$\Rightarrow 4462000cm^3$

In this problem, the least number of significant figures is 4.

So, the correct answer is, $4.462\times 10^{6}cm^3$

Hence, the number of significant figures are, four (4).

Yuri [45]2 years ago

4 0

4 - due to being a multiplication problem the answer goes to the least amount of Sig Figs given in a previous number.

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How many moles of methane (CH4) could be made from 4.6 moles of hydrogen?

DENIUS [597]

Answer:

Number of moles of methane form = 2.3 mol

Explanation:

Given data:

Number of moles of Hydrogen = 4.6 mol

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2 years ago

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2 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

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2 years ago

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Gnesinka [82]

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2 years ago

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3.0 L of oxygen gas is heated from 100k to 250k. What is the new volume after the temperature has been increased???

alexandr402 [8]

Answer:

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Explanation:

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2 years ago