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:
<h3>The answer is 6.93 cm³</h3>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question we have
We have the final answer as
<h3>6.93 cm³</h3>
Hope this helps you
The image of the bonds are missing, so i have attached it.
Answer:
A) - Sigma bond
-Sp³ and Sp³
- None
B) - Sigma and pi bond
- Sp² of C and p of O
- p of C and P of O
Explanation:
A) For compound 1;
- the molecular orbital type is sigma bond due to the end-to-end overlapping.
- Atomic orbitals in the sigma bond will be Sp³ and Sp³
- Atomic orbitals in the pi bond would be nil because there is no pi bond.
B) For compound 2;
- the molecular orbital type is sigma and pi bond
-Atomic orbitals in the sigma bond would be Sp² of C and p of O
- The Atomic orbitals in the pi bond will be; p of C and p of O
I think the correct answer is B. J.J. Thompson's Plum Pudding model had to be revised because atoms have protons and neutrons in the nucleus. He realized that aside from the electrons which are negatively charged, positive charge particles should be present to neutralize the charge of an atom.
