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2 years ago

What do potatoes, wood, and lobster shells have in common?

1 answer:

5 0

They are all biotic factors, meaning they were once alive or were alive. (an abiotic factor is something that has never lived.) hope this helped.

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A compound containing Na, C, and O is found to have 1.06 mol Na, 0.528 mol C, and 1.59 mol O. What is the empirical formula of t

Effectus [21]

Given :

Moles of Na : 1.06

Moles of C : 0.528

Moles of O : 1.59

To Find :

The empirical formula of the compound.

Solution :

Dividing moles of each atom with the smallest one i.e 0.528 .

So,

Na : 1.06/0.528 = 2.007 ≈ 2

C : 0.528/0.528 = 1

O : 1.59/0.528 = 3.011 ≈ 3

Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.

So, empirical formula is $Na_2CO_3$ .

Hence, this is the required solution.

5 0

2 years ago

In the following reaction, how many grams of silver can be produced from 49.1 g copper?

notsponge [240]

166.4 g Ag grams of silver can be produced from 49.1 g of copper.

A mole is a very important unit of measurement that chemists use. A mole of something means you have 602,214,076,000,000,000,000,000 of that thing, like how having a dozen eggs means you have twelve eggs.

$1 mol Cu + 2 mol AgNO_3$ → $2 mol Ag + 1 mol Cu(NO3)2$

63.55 g Cu —> 2 x 107.688 g Ag

63.55 g Cu gives 215.376 g of Ag

So, 49.1 g Cu —> $\frac{215.376 g X 49.1 g}{63.55 g}$

= 166.4 g Ag

Hence, 166.4 g Ag grams of silver can be produced from 49.1 g of copper.

Learn more about moles here:

brainly.com/question/26416088

#SPJ1

5 0

1 year ago

20 PTS!!

uranmaximum [27]

The correct answer is C because a Controlled Experiment tests only one factor at a time and consists of a control group and one or more experimental groups.

6 0

8 months ago

Determine the mass of Al(C2H3O2)3 that contains 2.63 Χ 1024 atoms of oxygen.

sergejj [24]

The molecular formula of compound is $Al(C_{2}H_{3}O_{2})_{3}$ .

Since, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6.023\times 10^{23}$ atoms of $Al(C_{2}H_{3}O_{2})_{3}$ .

Thus, according to molecular formula, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6\times 6.023\times 10^{23}=3.61\times 10^{24}$ atoms of oxygen atoms.

1 atom of oxygen will be present in $\frac{1}{3.61\times 10^{24}}$ moles of $Al(C_{2}H_{3}O_{2})_{3}$ . Thus,

$2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}$ .

Molar mass of $Al(C_{2}H_{3}O_{2})_{3}$ is 236 g/mol, mass can be calculated as follows:

$m=n\times M=0.7285 mol\times 236 g/mol=172 g$

Therefore, mass of $Al(C_{2}H_{3}O_{2})_{3}$ will be 172 g.

5 0

2 years ago

Read 2 more answers

For the unbalanced equation, what is the coefficient for c5h10 ?

marshall27 [118]

The answer is the coefficient is "1".
C₅H₁₀, now you see that there is no number with this and when there is no number or digit, it means the coefficient is one.
we use the coefficients to balance the equation of the reaction in such a way that number of the atoms of the elements in the reactants are equal to the number of atoms of different elements in the product, so that both sides are equal and balanced.

8 0

2 years ago