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3 years ago

Determine the mass in grams of 0.600 mol of oxygen atoms ( 32 grams x1mole/(mass of element)

Chemistry

1 answer:

yarga [219]3 years ago

8 0

Answer:

The mass of 02 is 19,2 grams in 0,6 moles

Explanation:

1 mol 0 2------------32g

0,6mol 02-----------x= (0,6 mol 02 x 32 g)/1 mol 02 = 19,2 grams

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What year will Aquarius be visible from Earth’s North Pole?

Minchanka [31]

Answer:

Aquarius beginning in the mid-3rd millennium. The north and south celestial poles are the two imaginary points in the sky where the Earth's ... In about 5,500 years, the pole will have moved near the position of the star ... The south celestial pole is visible only from the Southern Hemisphere.

Explanation:

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3 years ago

What mass of solid that has a molar mass of 46.0 g/mol should be added to 150.0 g of benzene to raise the boiling point of benze

enyata [817]

Answer : 17.12 g

Explanation: $\Delta T =k_b\times m$

$\Delta T$ = elevation in boiling point

$k_b$ = boiling point elevation constant

m= molality

$molality=\frac{mass of solute}{molecular mass of solute\times weight of the solvent in kg}$

given $\Delta T=6.28^{\circ}C$

Molar mass of solute = 46.0 $gmol^{-1}$

Weight of the solvent = 150.0 g = 0.15 kg

Putting in the values

$molality=\frac{x}{46gmol^{-1}\times0.15kg}$

$6.28 =2.53^{\circ}Ckgmol^{-1}\frac{x}{46gmol^{-1}\times 0.15kg}$

x = 17.12 g

3 0

3 years ago

I will give brainliest!!! In the reaction represented by the equation Al2O3 -> Al + O2, what is the mole ratio of aluminum to

Effectus [21]

Answer:;

4:3

Explanation:

Balanced:

4Al + 302 yields 2al2O3

7 0

3 years ago

Given the balanced equation what is the reaction ?

VMariaS [17]

This is a double replacement reaction; the ions switch twice.

3 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago