Answer:
A. Increasing the temperature will favor forward reaction and more CaCo3 formed.
B. More CaCo3 will be formed.
C. CaCo3 will decrease and more react ants formed.
D. Less CaCo3 will be formed.
E. Iridium is a catalyst so there is no effect
Explanation:
A. Temperature will increase because it's an endothermic reaction.
B. Adding Cao will favor forward reaction and more CaCo3 formed.
C. Removing methane, more react ants are formed and CaCo3 decreases.
D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.
The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.
Isotope mass amu Relative abundance
1 77.9 14.4
2 81.9 14.3
3 85.9 71.3
Express your answer to three significant figures and include the appropriate units.
Answer: 84.2 amu
Explanation:
Mass of isotope 1 = 77.9
% abundance of isotope 1 = 14.4% = 
Mass of isotope 2 = 81.9
% abundance of isotope 2 = 14.3% = 
Mass of isotope 3 = 85.9
% abundance of isotope 2 = 71.3% = 
Formula used for average atomic mass of an element :

![A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2877.9%5Ctimes%200.144%29%2B%2881.9%5Ctimes%200.143%29%2B%2885.9%5Ctimes%200.713%29%5D)

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu
Answer:
pH = 7.08
Explanation:
HCl ---------> H^+ + Cl^-
It's an acid, we are using this formula
pH = -log [H]^+
H^+ = 8.4 * 10^-8
pH = - log [8.4 * 10^-8]
It can also be solved as
-log 8.4-(-8log 10)
-0.924-(-8×1)
-0.924+8
7.076
To the nearest hundredth
pH = 7.08
The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).
The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.
The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.
To view more about rational reaction, refer to:
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