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andriy [413]
3 years ago
12

10. A sample of an unknown composition was tested in a laboratory. The sample could not be broken down by

Chemistry
2 answers:
damaskus [11]3 years ago
8 0

Answer:

it would be an element because its an element

Explanation:

vfiekz [6]3 years ago
7 0
It’s element it’s correct
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Which of the following is not an example of a science?
Cloud [144]
Geology is the science of Earth.

Ecology is the science of relationship between an organism and its environment.

Astronomy is the science of studying astronomical objects.

Whereas, Phrenology is the study of relating the size of human skull like its size, shapes e.t.c to his/her personality. So, its not a science as there is no scientific relationship between it.
6 0
3 years ago
Given that the molar mass of NaNO3 is 85.00 g/mol, what mass of NaNO3 is needed to make 4.50 L of a 1.50 M NaNO3 solution?
marissa [1.9K]
Mass = molarity x molar mass( NaNO₃) x volume

mass = 1.50 x 85.00 x 4.50

mass = 573.75 g of NaNO₃

hope this helps!
8 0
3 years ago
Read 2 more answers
How many grams of NH3 can be produced from 12.0g of H2?
RSB [31]

Answer:

Balanced reaction:

3 H2 (g)  + N2 (g)  → 2 NH3 (g)

Use stoichiometry to convert g of H2 to g of NH3.  The process would be:

g H2 → mol H2 → mol NH3 → g NH3

12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3

Explanation: See above

Hope this helps, friend.

8 0
1 year ago
Which of these is NOT a property of water?
Elis [28]

Answer:

-low specific heat

3 0
3 years ago
Read 2 more answers
8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a
Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

C_{1} V_{1} =C_{2} V_{2}

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, (5.45*V_{1} ) = (820.7*1.41)

V_{1}=\frac{820.7*1.41}{5.45}=  212.33 ml

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

3 0
2 years ago
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