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2 years ago

11

find the weight of the same object on a planet where the gravitational attraction has been reduced to 1/10 of the earth pull sho

w all work

1 answer:

nataly862011 [7]2 years ago

6 0

Answer:

9.9Xkgms-¹

Explanation:

Let mass of object be xkg

Gravitational pull is 10m/s-1/10=99/10=9.9

9.9xkgms-¹

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A reaction produces 10.5 L if oxygen, but was supposed to produce 1 mol of oxygen. What is the percent yield ? (One mole of any

Rzqust [24]

To answer this item, we assume that oxygen behaves ideally such that it is able to fulfill the following equation,

PV = nRT

If we are to retain constant the variable n and V.

The percent yield can therefore be solved through the following calculation,

n = (10.5 L)/(22.4 L) x 100%

Simplifying,
n = 46.875%

Answer: 48.87%

5 0

2 years ago

a rock dropped in a graduated cylinder raises the level of water from 20 ml to 35 ml. the rock has mass of 45 g. what is the den

adelina 88 [10]

Answer:

15...

Explanation:

6 0

2 years ago

Someone please do this!!!

Lilit [14]

Answer:

1) Fe = 69.9%

O = 31.1%

2) H = 5.19%

O = 16.5%

N = 28.9%

C = 49.5%

Explanation:

One easy way to do percent compositions is to assume you have 100g of a substance.

1) Lets say we have 100g of Fe2O3.

The total molar mass would be:

$= 55.845*2+15.999*3 = 159.687$

The molar mass of the Fe2 alone is:

$=55.845*2 = 111.69$

Thus, the grams of Fe2(out of a 100) could be calculated by multiplying 100g * the molar mass ratio of Fe2 to the whole:

$= 100g *\frac{111.69}{159.687} = 69.9431$

Which is approximately 69.9%.

We can find the amount of O3 by simply subtracting, as the rest of the compound is made of O3. Thus, the % composition of O3 is 31.1%

You can then do this same process to the next question, getting us the following:

H = 5.19%

O = 16.5%

N = 28.9%

C = 49.5%

4 0

2 years ago

Does anyone know how to do this? the solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What i

g100num [7]

It's simple, just follow my steps.

1º - in 1 L we have $0.0100~g$ of $BaCO_3$

2º - let's find the number of moles.

$\eta=\frac{m}{MM}$

$\eta=\frac{0.0100}{197.3}$

$\boxed{\boxed{\eta=5.07\times10^{-5}~mol}}$

3º - The concentration will be

$C=5.07\times10^{-5}~mol/L$

But we have this reaction

$BaCO_3\rightleftharpoons Ba^{2+}+CO_3^{2-}$

This concentration will be the concentration of $Ba^{2+}~~and~~CO_3^{2-}$

$K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}$

considering $[BaCO_3]=1~mol/L$

$K_{sp}=[Ba^{2+}][CO_3^{2-}]$

and

$[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L$

We can replace it

$K_{sp}=(5.07\times10^{-5})*(5.07\times10^{-5})$

$K_{sp}\approx25.70\times10^{-10}$

Therefore the $K_{sp}$ is:

$\boxed{\boxed{\boxed{K_{sp}\approx2.57\times10^{-11}}}}$

4 0

3 years ago

Read 2 more answers

How many formula units of sodium chloride (NaCl) may theoretically be produced from 13.0 g FeCl3?

algol [13]

Answer:

$1.445\times 10^{23}$ formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

Explanation:

Mass of ferric chloride = 13.0 g

Moles of ferric chloride = $\frac{13.0 g}{162.5 g/mol}=0.08 mol$

1 mole of ferric chloride has three moles of chloride ions.Then 0.08 moles of ferric chloride will have :

$3\times 0.08 mol=0.24 mol$ of chloride

$Na^++Cl^-\rightarrow NaCl$

1 mole of sodium ion reacts with 1 mole of chloride ion to form 1 mole of NaCl. Then 0.24 moles of chloride ion will give:

$\frac{1}{1}\times 0.24 mol=0.24 mol$ of NaCl

1 mole = $N_A=6.022\times 10^{23}$ molecules/ atoms

Number of NaCl molecules in 0.24 moles :

$=6.022\times 10^{23}\times 0.24=1.445\times 10^{23} molecules$

$1.445\times 10^{23}$ formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

4 0

3 years ago