Answer:
37.1°C.
Explanation:
- Firstly, we need to calculate the amount of heat (Q) released through this reaction:
<em>∵ ΔHsoln = Q/n</em>
no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.
<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>
∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.
Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = 2781.87 J).
m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).
c is the specific heat capacity of water (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).
∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)
∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.
<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>
Answer:
The answer to your question is 0.5 moles
Explanation:
Data
moles of Glucose = ?
moles of carbon dioxide = 3
Balanced chemical reaction
6CO₂ + 6H₂O ⇒ C₆H₁₂O₆ + 6O₂
Process
To solve this problem, use proportions, and cross multiplication.
Use the coefficients of the balanced equation.
6 moles of CO₂ ----------------- 1 mol of C₆H₁₂O₆
3 moles of CO₂ ---------------- x
x = (3 x 1) / 6
-Simplification
x = 3/6
-Result
x = 0.5 moles of Glucose
<span>For each inherited character an organism has two alleles for the gene controlling that character.. one form each parent</span>
The computation for molarity is:
(x) (0.175 L) = 0.0358 g / 598 g/mol
x = 0.000342093 M
Whereas the osmotic pressure calculation:
pi = iMRT
pi = (1) (0.000342093 mol/L) (0.08206 L atm / mol K) (298 K)
pi = 0.0083655 atm
Converting the answer to torr, will give us:
0.0083655 atm times (760 torr/atm) = 6.35778 torr
which rounds off to 6.36 torr
