What is the volume of a substance with a mass of 64.9 g/ and a density of 2.1 g/ml?

There are three subatomic particles found inside atoms. Two of the subatomic particle have a charge of positive or negative. The

NARA [144]

Neutrons don't have a charge. They are neutral

7 0

3 years ago

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Really need help with this! Chemistry

Westkost [7]

Answer:

a) 0,5

Explanation:

If x=6 and y=2, then (2x-4y)/(x+y)=(2*6-4*2)/(6+2)=(12-8)/8=4/8= 0,5

5 0

2 years ago

The ions Cr3+ and O2- combine to form the ionic compound chromium oxide. In what proportions do these ions combine to produce a

VMariaS [17]

<span>Cr3+ O2- ; the formula is Cr2O3 ( criss cross)</span>

7 0

3 years ago

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A cup of coffee has 71 mL of coffee and 127 mL of water. What is the percent volume of the coffee solution?

Anna [14]

Answer:

35.9%

Explanation:

The percent volume of the coffee solution can be calculated as follows:

% volume of coffee solution = volume of coffee/total volume of coffee solution × 100

According to this question, a cup of coffee has 71 mL of coffee and 127 mL of water. This means that, the total volume of coffee solution is;

71mL + 127mL = 198mL

% volume = 71/198 × 100

= 0.359 × 100

Percent volume of coffee solution = 35.9%

3 0

3 years ago

One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su

Tpy6a [65]

Answer : The percent purity of $Fe_2O_3$ in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

$Fe_2O3+3CO\rightarrow 2Fe+3CO_2$

First we have to calculate the mass of Fe.

$\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}$

Molar mass of Fe = 55.8 g/mole

$\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole$

Now we have to calculate the moles of $Fe_2O_3$

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of $Fe_2O_3$

So, $3.15\times 10^4mole$ of Fe produced from $\frac{3.15\times 10^4}{2}=15750$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3$

Molar mass of $Fe_2O_3$ = 159.69 g/mole

$\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg$

Now we have to calculate the percent purity of $Fe_2O_3$ in the original sample.

Mass of original sample = $2.86\times 10^3kg$

$\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100$

$\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%$

Therefore, the percent purity of $Fe_2O_3$ in the original sample is 87.94 %

3 0

3 years ago