<u>Given:</u>
Volume of the unknown monoprotic acid (HA) = 25 ml
<u>To determine: </u>
The concentration of the acid HA
<u>Explanation:</u>
The titration reaction can be represented as-
HA + NaOH → Na⁺A⁻ + H₂O
As per stoichiometry: 1 mole of HA reacts with 1 mole of NaOH
At equivalence point-
moles of HA = moles of NaOH
For a known concentration and volume of added NaOH we have:
moles of NaOH = M(NaOH) * V(NaOH)
Thus, the concentration of the unknown 25 ml (0.025 L) of HA would be-
Molarity of HA = moles of HA/Vol of HA
Molarity of HA = M(NaOH)*V(NaOH)/0.025 L
The answers here is B) Before, the substance was a gas, later it was a liquid.
Gas particles move freely and away from each other. However, liquid particles move around each other.
Hope this helps! :)(
<h3><u>Answer;</u></h3>
1. In the light reactions, light energy is used to oxidize H2O to O2.
2. The electrons derived from this oxidation reaction in the light reactions are used to reduce NADP+ to NADPH.
3. The Calvin cycle oxidizes the light-reactions product NADPH to NADP+.
4. The electrons derived from this oxidation reaction in the Calvin cycle are used to reduce CO2 to G3P.
<h3><u>Explanation;</u></h3>
- <em><u>In the light reactions, light energy is used to remove electrons from (oxidize) water, producing O2 gas. These electrons are ultimately used to reduce NADP+ to NADPH.
</u></em>
- In the Calvin cycle, NADPH is oxidized back to NADP+ (which returns to the light reactions). The electrons released by the oxidation of NADPH are used to reduce three molecules of CO2 to sugar (G3P), which then exits the Calvin cycle.
- As ATP and NADPH are used in the Clavin cycle, they produce ADP and NADP+, respectively, which are returned to the light reactions so that more ATP and NADPH can be formed.
Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.
2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O
1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O
Empirical formula
H2SO4
The question is incomplete, the complete question is;
Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75
Answer:
A nucleus with a A nucleus with a neutron:proton ratio of 1.49
A nucleus with a mass of 187 and an atomic number of 75
Explanation:
The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1
Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.
Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.
Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.
