Answer:
D. C14 appears at C‑6 of fructose 6‑phosphate
Explanation:
Conversion of glucose B-phosphate via 6-phospho-
gluconate to pentose phosphate and CO2 followed by conversion
of the pentose phosphate to fructose 6-phosphate and glyceralde-hyde phosphate followed by conversion of the fructose 6-phos-
phate to glucose g-phosphate .
0.1 mol/L
. The concentration of the HCl is 0.1 mol/L
a) Write the <em>balanced chemical equation
</em>
HCl + NaOH → NaCl + H2O
b) Calculate the <em>moles of NaOH
</em>
Moles of NaOH = 0.050 L NaOH x (0.1 mol NaOH/1 L NaOH)
= 0.0050 mol NaOH
c) Calculate the <em>moles of HCl
</em>
Moles of HCl = 0.0050 mol NaOH x (1 mol HCl/1 mol NaOH)
= 0.0050 mol HCl
d) Calculate the <em>molar concentration</em> of the HCl
<em>c</em> = moles/litres = 0.0050 mol/0.050 L = 0.1 mol/L
Answer:
B. Add more water to decrease the concentration.
Explanation:
If we add more water to the solution, the concentration is decreased and like wise the conductivity of the solution.
- Concentration is the amount of solute in a solution.
- Conductivity deals with the ability of a solution to allow the passage of current.
Concentration and conductivity are directly proportional to one another. The higher the concentration, the higher the conductivity and vice versa.
- Adding more water increases the amount of solvent and the concentration reduces.
<h3>
Answer:</h3>
4.73 × 10^4 m
<h3>
Explanation:</h3>
From the question;
Frequency of the photon = 634 × 10^12 Hz
We are required to calculate the wavelength of the photon.
We need to know the relationship between wavelength and frequency of a wave.
The relationship between f and λ is given by;
c = fλ
Where c, is the speed of light, 2.998 × 10^8 m/s
Therefore, to get the wavelength we rearrange the formula such that;
λ = c ÷ f
= 2.998 × 10^8 m/s ÷ 634 × 10^12 Hz
= 4.73 × 10^-5 m
But we require wavelength in nm
1 M = 10^9 nm
Therefore;
Wavelength = 4.73 × 10^-5 m × 10^9 nm/m
= 4.73 × 10^4 m
Hence, the photon's wavelength is 4.73 × 10^4 m
B. 1, 1, 1, 2
Explanation:
You only need to balance the NaNO3 on the right. Since there is 2 NO3 on the left, you need to put a 2 in front of the NaNO3 on the right. Everything else is already balanced so the only coefficient needed is 2 in front of the NaNO3.
