The density of He is 1.79 x 10⁻⁴ g/mL
In other words in 1 mL there's 1.79 x 10⁻⁴ g of He.
To fill a volume of 6.3 L the mass of He required
= 1.79 x 10⁻⁴ g/mL * 6300 mL
= 11 277 * 10⁻⁴ g
Therefore mass of He required = 1.1277 g of He
<h3><u>Answer</u>;</h3>
1.0875 x 10-2 atm
<h3><u>Explanation;</u></h3>
2O3(g) → 3O2(g)
rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t
The average rate of disappearance of ozone ... is found to
be 7.25 × 10–3 atm over a certain interval of time.
This means (ignoring time)
∆[O3]/∆t = -7.25 × 10^–3 atm
(it is disappearing, thus the negative sign)
rate = -(1/2)∆[O3]/∆t
rate = -(1/2)*(-7.25 × 10^–3 atm)
= 3.625 × 10^–3 atm
Now use the other part of the expression:
rate = +(1/3)∆[O2)∆t
3.625 × 10–3 atm = +(1/3)∆[O2)/t
∆[O2)/∆t = (3)*(3.625× 10^–3 atm)
= 1.0875 x 10-2 atm over the same time interval
Answer: HCl
Explanation:
calcium carbonate dissolves in HCl acid producing CO 2 gas. It will not dissolve in pure water. The Ksp for calcium carbonate in water is 3.4 x 10-9 moldm-3 which is very low. What takes place here is actually a chemical reaction:
CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + H 2CO 3(aq)
This reaction accounts for the solubility of the Calcium carbonate in HCl and not in pure water.
Answer:
Use the formula q = m·ΔHv in which q = heat energy, m = mass, and ΔHv = heat of vaporization.
Explanation:
:)
Answer:
The claim is: Therefore Mitosis requires less energy than sexual reproduction does.
Explanation:
