A because that honestly just makes the most sense
Answer:
0.1g (Gallon) of chlorine
Explanation:
<u>Formula</u>
1 gallon = 3.7L; the density of water is 1.0g/ml
<u>Given</u>
2g (gallon) of chlorine to sanitize = 1,000,000g (gallon) of water
<u>Solve</u>
If 2g (gallon) chlorine = 1,000,000g (gallon)
∴, ? chlorine = 40,000
The First step; set up an equation
1000000/2 = 40000/?
The Next step; divide 1 million to 2
1000000 ÷ 2 = 500000
Then, divide the result by 40000
40000 ÷ 500000 = 0.08
In the nearest unit that is 0.1
Therefore, it will take 0.1g (gallon) of chlorine to sanitize a 40,000-gallon pool.
Given:
Q = 9.4 kJ/(kg-h), the heat production rate
c = 4.18 J/(g-K), the heat capacity
t = 2.5 h, amount of time
Note that
c = 4.18 J/(g-K) = 4180 J/(kg-K) = 4.18 kJ/kg-K)
Consider 1 kg of mass.
Then
Qt = cΔT
where ΔT is the increase in temperature (°K)
(1 kg)*(9.4 kJ/(kg-h))*(2.5 h) = (1 kg)*(4.18 kJ/(kg-K))*(ΔT K)
23.5 = 4.18 ΔT
ΔT = 23.5/4.18 = 5.622 K = 5.622 °C
Answer: 5.62 K (or 5.62 °C)
Answer:
Mass of NaBr produced = 23.67 g
Explanation:
Given data:
Mass of AgBr = 42.7 g
Mass of NaBr produced = ?
Solution:
Chemical equation:
2Na₂S₂O₃ + AgBr → NaBr + Na₃(Ag(S₂O₃)₂
Number of moles of AgBr:
Number of moles = mass/molar mass
Number of moles = 42.7 g/ 187.7 g/mol
Number of moles = 0.23 mol
now we will compare the moles of AgBr with NaBr.
AgBr : NaBr
1 : 1
0.23 : 0.23
Mass of NaBr:
Mass = number of moles × molar mass
Mass = 0.23 mol × 102.89 g/mol
Mass = 23.67 g
Answer:
pH = 12.08
Explanation:
First we <u>calculate how many moles of each substance were added</u>, using <em>the given volume and concentration</em>:
- HBr ⇒ 0.05 M * 75 mL = 3.75 mmol HBr
- KOH ⇒ 0.075 M * 74 mL = 5.55 mmol KOH
As HBr is a strong acid, it dissociates completely into H⁺ and Br⁻ species. Conversely, KOH dissociates completely into OH⁻ and K⁺ species.
As there are more OH⁻ moles than H⁺ moles (5.55 vs 3.75), we <u>calculate how many OH⁻ moles remain after the reaction</u>:
- 5.55 - 3.75 = 1.8 mmoles OH⁻
With that<em> number of moles and the volume of the mixture</em>, we <u>calculate [OH⁻]</u>:
- [OH⁻] = 1.8 mmol / (75 mL + 74 mL) = 0.0121 M
With [OH⁻], we <u>calculate the pOH</u>:
With the pOH, we <u>calculate the pH</u>:
