Answer:
The correct answer is Dust
Explanation
Dust is a dry dirt in powder form usually found on surfaces of items in a building, it comprises of very small particles of soil, sand and sometimes includes toxic substances, skin cells,bacteria, soil particles, particles of clothing material, tiny pieces of dead insects and pollen
Answer:
b) Phosphorus acid
Explanation:
To distinguish the type of acid of phosphorus with the oxidation state of +3, we need to be familiar with the chemical formula of each of the compounds:
Orthophosphoric acid H₃PO₄
Phosphorus acid H₃PO₃
Metaphosphoric acid HPO₃
Phyrophosphoric acid H₄P₂O₇
Now that we know the formula of the given compounds, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero:
Only phosphorus acid yielded an oxidation state of +3 for phosphorus in the compound.
H₃PO₃:
we know the oxidation state of H = +1
O = -2
The oxidation state of P is unknown. We can express this as an equation:
3(+1) + P + 3(-2) = 0
3 + P -6 = 0
P-3 = 0
P = +3
+
⇔
Decreasing the temperature of the reaction,the reaction shifts forward.
The explanation is given below.
Explanation:
If the temperature of the reaction mixture is increased,then the equilibrium will shift to decrease the temperature.
If the temperature of the reaction mixture is decreased,then the equilibrium will shift to increase the temperature.
During the formation of the ammonia,it gives off heat.So it is an exothermic reaction.
+⇔
A decrease in the temperature favors the reaction that is exothermic (the forward reaction)because it produces energy.Therefore,if the temperature is decreased,the yield of the ammonia increases.
<em>Therefore if the temperature is increased,the reaction shifts forward and the yield of the ammonia increases and it is an exothermic reaction.</em>
Answer:
CH2O
Explanation:
Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.
C = 5.692/14.229 * 100 = 40%
O = 7.582/14.229 * 100 = 53.29%
H = 0.955/14.229 * 100 = 6.71%
We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.
C = 40/12 = 3.333
O = 53.29/16 = 3.33
H = 6.71/2 = 6.71
Dividing by the smaller value which is 3.33
C = 3.33/3.33 = 1
O = 3.33/3.33= 1
H = 6.71/3.33 = 2
The empirical formula of the compound ribose is CH2O
