The volume of 1.0 g of nitrogen in a balloon is increased. It causes: the mass of the gas to decrease the density of the gas to increase and the pressure of the gas to decrease the temperature of the gas decrease.
<h3>What is volume?</h3>
How much space an object or substance takes up.
If you increase the temperature of the gas in the balloon, the pressure will initially increase. The balloon will expand until the pressure inside drops back down and is equal to the pressure outside.
If you increase the atmospheric pressure, the balloon will compress until the pressure inside equals the new pressure outside.
If you remove gas from the balloon, the pressure will drop, causing the balloon to compress until the pressure is again equal to the pressure outside.
if you place the balloon underwater, it will be under greater pressure, causing the balloon to compress until the inside pressure equals the new outside pressure.
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The answer here would be 14.9 %KCl and here is how I can explain why:
33.5g / 225.6g x 100% = 14.9%
<span>you are looking for % of KCl in the solution, you have to add the mass of the KCl and water to get the total mass of the solution
Hope this helps a lot
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Is there something else that goes along with this?
Answer:
The answer to your question is: [H+] = 10⁻¹²
Explanation:
Remember that pH = -log [H+], then, in this exercise we need to calculate the pH of the substances and compare the results, the one which the higher value will be the strongest base.
a) [H+] = 10^-9 pH = -log[10⁻⁹] = 8
b) [H+] = 10^-11 pH = -log[10⁻¹¹] = 10
c) [H+] = 10^-10 pH = -log[10⁻¹⁰] = 9
d) [H+] = 10^-12 pH = -log[10⁻¹²] = 11 This is the strongest base
Answer: PH= 1.4
Explanation:
n(mole)= concentration(c)× volume(v)
n(HClO4) = 0.08 ×0.03
n(HClO4)= 0.0024 mole
n(Ba(OH)2) = 0.28 × 0.05
n(Ba(OH)2) = 0.014
2HClO4 + Ba(OH)2.....>Ba(ClO4)2 + 2H2O.
Hence, the limiting reagent is perchloric acid. According to the reaction:
0.0024 HClO4 req 1/2× 0.0024 of Ba(OH)2.
n(Ba(OH) reacted= 0.0012
Excess n(Ba(OH)2) = 0.014-0.0012
Excess n(Ba(OH)2) = 0.0128
Conc(Ba(OH)2) unreacted = 0.0128/.32 (vol. of mix is 32ml)
= 0.04M = [H+]
PH = -log[0.04] = 1.397
PH= 1.4
