Answer:
first let's find the molar mass of CH2.Molar mass CH2=12.0+(2multiply1.01)=14.02g/mol
Explanation:
A. 3<span> chemical reactions are possible regarding the formation of carbonic acid and the dissolving of calcite.</span>
Answer: m Most likely Group2 metals and Group 17 non-ktta
Explanation:
A little more information is needed to be certain, but the likely answer is that X belongs to Group 2 and Y belongs to Group 17. Group 2 metals (Be, Mg, Ca, Sr, Ba, etc.) are all divalent. They gave rive up 2 electrons each to return to a full shell. Group 17 elements (e,g, F, Cl, Br, I, etc.) all require 1 electron to reach a full valence shell. That would make the proportion 1X to 2Y, or XY2. It is possible that a metal outside of Group 2 would also have a valency of 2. Iron(II) forms FeCl2, for example.
Answer : The value of 
for the final reaction is, 184.09
Explanation :
The equilibrium reactions in aqueous solution are :
(1) 
(2) 
The final equilibrium reaction is :
Now we have to calculate the value of for the final reaction.
Now equation 1 is multiply by 2 and reverse the equation 2, we get the value of final equilibrium reaction and the expression of final equilibrium constant is:
Now put all the given values in this expression, we get :
Therefore, the value of for the final reaction is, 184.09
<span>1.91 mol H2
Let's rewrite that equation into a balanced equation. Let's start with the unbalanced version:
Na + HF ==> NaF + H2
We have equal amounts of Na and F on both sides, but twice as much H on the right. So let's double the Na, HF, and NaF coefficients.
2Na + 2HF ==> 2NaF + H2
And we now have equal quantities of all elements on both sides, so our equation is balanced.
Now looking at the balanced equation, 2 moles of sodium should produce 1 mole of hydrogen gas. So we should have 4.50/2 = 2.25 moles of hydrogen gas. But we only have an 85% yield, so we multiply the expected yield by the percentage, giving 2.25 mol * 0.85 = 1.9125 mol. Rounding to 3 significant figures gives us 1.91 mol which matches "1.91 mol H2"</span>
