The choices for this problem are bismuth, Bi; platinum, Pt; selenium, Se; calcium, Ca and copper, Cu. I think the correct answer would be selenium. The melting point of bismuth is at a temperature of 544.4 Kelvin. At a temperature of 525 K, it would exist as solid. Platinum melts at 2041.1 K. At 525 K, platinum would be in solid form. Selenium has a melting point at 494 K so that at a temperature of 525 K, it would exist in its liquid state. Calcium has a melting point of 1112 K so it would exist as solid at 525 K. Copper has a melting point at 1358 K, so it would still exist as solid at a temperature of 525 K. Therefore, the answer would only be selenium.
Reaction involved in present electrochemical cell,
At Anode: Zn → Zn^2+ + 2e^2-
At cathode: Zn^2+ + 2e^2- → Zn
Net Reaction: Zn + Zn^2+ ('x' m) → Zn^2+(0.1 m) + Zn
Number of electrons involved in present electrochemical cell = n = 2
According to Nernst equation for electrochemical cell,
Ecell = -2.303
= 0.014
Given: T =
, F = 96500 C, R = gas constant = 8.314J/K.mol, [Zn^2+]R = 0.1 m , Ecell = 0.014 v
∴ 0.014 = - 2.303
∴ log
=
= -2.1117
∴ log x = log(0.1) + 2.1117
∴x = 13.09 m
The balanced chemical
reaction will be:
CH4 + 2O2 → CO2 + 2H2O
We are given the amount of carbon dioxide to produce from the reaction.
This will be our starting point.
560 L CH4 ( 1 mol CH4/ 22.4 L CH4 ) (2 mol O2/ 1 mol CH4 ) (
22.4 L O2 / 1 mol <span>O2</span><span>) = 1120 L O2</span>
Answer:
12 mmilligrams of Po-218 was the mass of the original starting material
Explanation:
The half-life of a radioactive material is the time taken for half the amount ofnthe original material present in a radioactive material to decay or disintegrate.
After each half-life, half the original material present at the start remains.
For the radioactive polonium-218 having a half-life of 3.04 minutes, it means that if 1 g is the starting material, after 3.04 minutes, 1/2 g will be remaining; after, 6.08 minutes 1/2 of 1/2 which is 1/4 of the starting material will be remaining; and after 9.12 minutes, 1/2 of 1/4 = 1/8 g will be remaining.
From the question, number of half-lives undergone after 9.12 minutes = 9.12/3.04 = 3 half-lives.
After 3 half-lives, 1/8 of the original material is remaining.
1/8 = 1.50 mg
The original mass of the sample at the start = 1.50 mg × 8 = 12 mg
Therefore, 12 milligrams of Po-218 was the mass of the original starting material.
The product will not be affected by the addition of twice as much Na₂CO₃.
<h3>What is Limiting reagent in stoichiometry ?</h3>
- The maximum quantity of the end product determined by a balanced chemical equation is known as the Stoichiometry.
- The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced, and the one which remains unconsumed after the final reaction is in Excess.
- Calculate the moles of each reactant present and contrast it with the mole ratio of the reactants in the balanced equation to determine which reactant is the limiting one.
Here,taking the stoichiometry into consideration, we find that the reaction happens with 1:1 ratio; so, adding twice the amount of Na₂CO₃ will lead to its excess making the other the limiting reactant, hence, it would not affect the yield of the product.
To know more about the Limiting reactant, refer to:
brainly.com/question/14222359
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