Answer:
OPTION (A) : Testing a rock sample for gold content
Explanation:
For testing a rock sample of gold content you will need a Chemist. To test the material, the sample is rubbed on black stone which will leave a mark on the stone. This mark is tested by applying aqua fortis i.e nitric acid on the mark. If the mark gets dissolve then the material is not gold. If the mark sustain the it is further tested by applying aqua regia i.e nitric acid and hydrochloric acid which will prove the sample is of gold if it gets dissolve on using hydrochloric acid. The purity of the sample can be checked by differing the concentration of the aqua regia and comparing it with the gold material of the known purity.
<span> 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
from the reaction 2 mol 4 mol
from the problem 5.4 mol 10.8 mol
M(CO2) = 12.0 +2*16.0 = 44.0 g/mol
10.8 mol CO2 * 44.0 g CO2/1 mol CO2 = 475.2 g CO2 </span>≈480 = 4.8 * 10² g
Answer is C. 4.8*10² g.
Answers are:
Catabolism:
- g<span>enerally exergonic (spontaneous): In this reactions energy is released.
- </span><span>convert NAD+ to NADH. Electrons and protons released in reactions are attached to NAD+.
- </span><span>generation of ATP. ATP is synthesis from ADP.
- </span><span>convert large compounds to smaller compounds. Foe example starch to monosaccaharides.
Anabolism:
</span><span>- convert NADPH to NADP+. Protons and electrons are used to make chemical bonds.
</span>- <span>convert small compounds to larger compounds.</span>
The balanced chemical reaction is written as:
<span>NaOH + HCl → NaCl + H2O
We are given the amount of sodium hydroxide to be used up in the reaction. This will be the starting point for the calculation.
2.75 x 10^-4 mol NaOH ( 1 mol H2O / 1 mol NaOH ) ( 18.02 g H2O / 1 mol H2O ) = 4.96 x 10^-3 g H2O</span>
