I just answered your other question and wrote all the definitions of the terms. I would say this is probably refering to Fusion because you are fusing together more nuclei into one nucleus. Fusion= bringing separate things into one entity.
Answer:
1.43 (w/w %)
Explanation:
HCl reacts with NH3 as follows:
HCl + NH3 → NH4+ + Cl-
<em>1 mole of HCl reacts per mole of ammonia.</em>
Mass of NH3 is obtained as follows:
<em>Moles HCl:</em>
0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>
<em>Mass NH3 in the aliquot:</em>
3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.
Mass of sample + water = 22.225g + 75.815g = 98.04g
Dilution factor: 98.04g / 14.842g = 6.6056
That means mass of NH3 in the sample is:
0.0545g * 6.6056 = 0.36g NH3
Weight percent is:
0.36g NH3 / 25.225g * 100
<h3>1.43 (w/w %)</h3>
Answer : The correct option is, (c) 79.62
Explanation :
The formula used for percent humidity is:
..........(1)
The formula used for relative humidity is:
...........(2)
where,
= partial pressure of water vapor
= vapor pressure of water
p = total pressure
First we have to calculate the partial pressure of water vapor by using equation 2.
Given:
Relative humidity = 80 % = 0.80
Now put all the given values in equation 2, we get:
Now we have to calculate the percent humidity by using equation 1.
Therefore, the percent humidity is 79.62 %
Answer:
The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²
Explanation:
The given parameters is as follows;
The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g
The required information = The number of formula units of potassium chloride (KCl)
The Molar Mass of KCl = 74.5513 g/mol
Therefore, we have;
1 mole of a substance, contains Avogadro's number (6.022 × 10²³) of formula units
Therefore;
0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³ ≈ 3.077588 × 10²² formula units
From which we have, the number of formula units in 3.81 g of potassium chloride (KCl) ≈ 3.08 × 10²² formula units.
Answer:
2
Explanation:
There are 3 carbons on the right side
there are only two on the left side....you need one more ...so add one more... change the '1' coefficient in front of C to '2'