Answer:
If the question is which can make a buffer, then NH3, NH4Cl should be correct. Because Ammonium (NH4) is conjugate acid of NH3 so they can form an equilibrium which is basically a buffer whose purpose is to resist pH change.
Explanation:
Answer:
315.
Explanation:
Hello.
In this case, since the given number has five significant figures as the zero is to the right of the first nonzero digit (3), if it is required to report it with three significant figures, it is necessary to "cut" it at the first five without any rounding since the subsequent zero is less than five.
Thus the number turns out:
315
Best regards.
1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol
</span>12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO
<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover.
CO + 2H2 -------> CH3OH
1 mol 2 mol
given 2.66 mol 6 mol (excess)
How much
we need CO? 3 mol 6 mol
We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.
3) </span>CO + 2H2 -------> CH3OH
1 mol 1 mol
2.66 mol 2.66 mol
4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol
2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
Answer:
See solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:
Thus, we solve for the molar mass of the metal to obtain:
For the subsequent problems, we proceed as follows:
a.
b.
c.
Regards!
<span>These atoms are known as valence atoms.</span>
