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4 years ago

What is the volume of 8.50 moles of hydrogen gas?

1 answer:

3 0

Okay!
8.50molH, represents 8.50 moles of Hydrogen gas!

To convert from moles to volume multiply it by the ratio of L/molH
mol of anything in a conversion problem is always 1 unless you are going to an unknown! Try to memorize this, but the number you always put in for the # of liters in a conversion problem is 22.4L (L is a measure of volume)

8.50molH * (22.4 L/ 1 molH)

190.L of Hydrogen Gas.

Hope that helped!

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Are any substance made up of matter and can be natural or man-made.

Mila [183]

Answer:

Yes.

Explanation:

Substances are made up of matter or matter are made up of tiny molecules or atoms that occur naturally or some are synthetic or man made.

All matter are made up of substances called elements and each elements have its own physical and chemical properties and cannot be broken easily by ordinary chemical reactions.

Of all the 118 elements only 92 occur naturally and 26 are man made or synthetic which are made in the laboratories.

3 0

3 years ago

What group on the Periodic Table contains the elements of the alkaline<br> earth family?

Jlenok [28]

Answer:

The alkaline earth metals are six chemical elements in group 2 of the periodic table. They are beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra).

Explanation:

hope that helps

5 0

3 years ago

For the reaction of 2Na2O, if you have 18.0 g of Na, how many grams of O2 are needed for reaction?

geniusboy [140]

Answer: -

6.26 grams

Explanation: -

Atomic Mass of sodium Na = 23 gram

Mass of Na = 18 g

Atomic mass of oxygen O = 16 gram

Molar mass of O2 = 16 x 2 = 32 gram

The balanced chemical equation for the reaction is

4 Na + O2 -- > 2 Na 2 O

From the balanced chemical equation, we see

4 Na requires 1 O2

4 x 23 gram of Na requires 32 gram of O2

18 g of Na gives $\frac{32 g O 2 x 18 g Na}{4 x 23 gram Na}$

= 6.26 grams of O2

For the reaction of formation of Na2O, for 18.0 g of Na, 6.26 grams of O2 are needed for reaction.

8 0

3 years ago

When 1.365 g of anthracene, C14H10, is combusted in a bomb calorimeter that has a water jacket containing 500.0 g of water, the

luda_lava [24]

<h3>The enthalpy of combustion per mole of anthracene : 7064 kj/mol(- sign=exothermic)</h3><h3>Further explanation </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

Heat released by anthracene= Heat absorbed by water

Heat absorbed by water =

$\tt Q=500\times 4.18\times 25.89=54110.1~J$

mol of anthracene (MW=178,23 g/mol)

$\tt \dfrac{1.365}{178.23}=0.00766$

The enthalpy of combustion per mole of anthracene :

$\tt \Delta H=-\dfrac{Q}{n}=\dfrac{54110.1}{0.00766}=-7063981.7~J/mol\approx -7064~kJ/mol$

8 0

3 years ago

Help help help help

Arada [10]

#3 would be 0.0056

#2 would be 9.92

#1 would be 0.011

8 0

3 years ago