It is called a waxxing gibbous, pls brainliest
Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
Answer:
The volume will be occupied is 244, 36L.
Explanation:
We convert the unit of temperature to celsius into Kelvin, then use the ideal gas formula, solve for V (volume) and use the gas constant R =0.082 l atm / K mol:
0°C=273K 25°C= 273 + 25=298K
PV=nRT ---> V=nRT/P
V= 5,00 mol x 0,082 l atm/ K mol x 298 K/0,500 atm
<em>V=244,36L</em>
Answer:
Na₂CO₃•H₂O
Explanation:
After it is heated, the remaining mass is the mass of sodium carbonate.
30.2 g Na₂CO₃
Mass is conserved, so the difference is the mass of the water:
35.4 g − 30.2 g = 5.2 g H₂O
Convert masses to moles:
30.2 g Na₂CO₃ × (1 mol Na₂CO₃ / 106 g Na₂CO₃) = 0.285 mol Na₂CO₃
5.2 g H₂O × (1 mol H₂O / 18.0 g H₂O) = 0.289 mol H₂O
Normalize by dividing by the smallest:
0.285 / 0.285 = 1.00 mol Na₂CO₃
0.289 / 0.285 = 1.01 mol H₂O
The ratio is approximately 1:1. So the formula of the hydrate is Na₂CO₃•H₂O.
