Answer:
28.52 L
Explanation:
First, let's calculate the density of the ocean, which is the mass divided by the volume:
d = m/V
d = 35.06/1
d = 35.06 g/L
So, for a mass of 1.00 kg = 1000.00 g
d = m/V
35.06 = 1000.00/V
V = 1000.00/35.06
V = 28.52 L
How all the data are expressed with two significant figures, the volume must also be expressed with two.
A conjugate acid<span>, within the </span>Brønsted–Lowry acid–base theory<span>, is a </span>species<span> formed by the </span>reception of a proton<span> (</span><span>H+</span><span>) by a </span>base<span>—in other words, it is a base with a </span>hydrogen<span> ion added to it. The conjugate acid would be as follows:
</span><span>Cl = HCl
CH2 = </span>H2O<span>
CO2- = </span><span>HCO3(1-)</span>
<span>The half-life of a first-order reaction is determined as follows:
</span>t½<span>=ln2/k
From the equation, we can calculate the </span><span>first-order rate constant:
</span>k = (ln(2)) / t½ = 0.693 / 90 = 7.7 × 10⁻³
When we know the value of k we can then calculate concentration with the equation:
A₀ = 2 g/100 mL
t = 2.5 h = 150min
A = A₀ × e^(-kt) =2 × e^(-7.7 × 10⁻³ × 150) = 0.63 g / 100ml
= 6.3 × 10⁻⁴ mg / 100ml
Answer:
The answer to your question is letter C
Explanation:
Data
Volume = 2 L
Molarity = 0.100 M
Molecular weight Na₂CO₃ = (2 x 23) + (1 x 12) + (3 x 16)
= 46 + 12 + 48
= 106 g
Process
1.- Calculate the grams of Na₂CO₃ needed
106 g ---------------- 1 mol
x ---------------- 0.1 moles
x = (0.1 x 106) / 1
x = 10.6 g
2.- Calculate the grams of Na₂CO₃ needed for 2 liters of solution
10.6 g -------------- 1 liter
x -------------- 2 liters
x = (10.6 x 2) / 1
x = 21.2 grams of Na₂CO₃
