The temperature change is 23 °C.
<em>q = mC</em>Δ<em>T</em>
Δ<em>T</em> = <em>q</em>/(<em>mC</em>)
<em>m</em> = 355 g
∴ Δ<em>T</em> = (34 000 J)/(355 g × 4.184 J·°C⁻¹g⁻¹) = 23 °C
<em>Note</em>: The answer can have only <em>two significant figures</em> because that is all you gave for the amount of heat absorbed.
As we know that Molarity is given as,
M = moles / V
Solving for V,
V = moles / M ------------------(1)
Also, moles is equal to,
moles = mass / M. mass -------------(2)
puting value of moles from eq. 2 into eq. 1,
V = (mass / M.mass) / M
Putting values,
V = (45 g / 164 g/mol) / 1.3 mol/dm³
V = 0.21 dm³
Answer:
Explanation:
Well use the combines gas law to evaluate this question;
P₁V₁/T₁ = P₂V₂/T₂ (Remember that at sea level the pressure is usually 1 atm)
1 * V₁ / 294.25 = 0.72 * V₂ / 277.95 (NB: Temperatures have to be in Kelvin)
V₁ / 294.25 * 277.95 = 0.72 V₂
0.945 V₁ = 0.72 V₂
0.945/0.72 V₁ = V₂
1.312 V₁ = V₂
The volume (V2) at 3000 m altitude will be <u>1.312 </u>bigger than the initial volume at sea level.
