Answer: The metal that is the cathode (B)
Explanation:
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<span>a decrease in Km and an increase in Vmax
km is the constant that relates how thick the volume of the substrate in terms of concentration when the speed is a large portion of the most extreme. They increment Km by meddling with the joining process that makes them stick together wit the substrate, however they have any effect with regards to Vmax since since the sticking process to ES did not occur</span>
<span>The
statements that are correct are ‘A large Keq value indicates that products are
favored.’ And ‘A small Keq value indicates that the reverse reaction will occur
very quickly.’ Having large Keq means that the concentration of the product is
greater than the reactants and so products are favored. Having small Keq means
that the concentration of the reactants is greater than the product.</span>
Answer:
Cools forming Igneous rock. Its weathered and eroded into sediments. The sediments are compacted and cemented forming sedimentary rock.
Explanation:
At STP, 1 mol of gas occupies 22.4 L of volume. Using this conversion we the starting amounts of the reactants:
1.6071 mol H2
.5357 mol CS2
The balanced equation is:
4H2 + CS2 -> CH4 + 2H2S
This means that 4 moles of H2 reacts with 1 mole of CS2.
Multiplying this ratio with each of the amounts of the reactants:
required CS2 based on amount of H2 = 0.4018 mol CS2
required H2 based on amount of CS2 = 2.1428 mol H2
Therefore, the excess reactant is CS2. To calculate how much is left, we simply subtract the amount of the reactant CS2 with the required amount based on the amount of H2:
0.5357 mol - 0.4018 mol = 0.1339 mol