SOMEONE PLS HELP ASAP IM DESPERATE

a 45g sample of water at 12°c is heated with 15kJ of energy. What will be the final temperature of the water?

Marizza181 [45]

Answer:

Final temperature = 91.75 °C

Explanation:

Given data:

Mass of water = 45 g

Initial temperature = 12 °C

Energy required = 15 Kj (15000 j)

Final temperature = ?

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Solution:

Q = m.c. ΔT

15000 j = 45 g × 4.18 j/g.°C × (T2 -T1)

15000 j = 45 g × 4.18 j/g.°C × (T2- 12 °C)

15000 j = 188.1 J/°C × (T2- 12 °C)

15000 j / 188.1 J/°C = T2- 12 °C

79.745 °C = T2 - 12 °C

T2 = 79.745 °C + 12 °C

T2 = 91.75 °C

4 0

3 years ago

How many electrons will each ele-

Andre45 [30]

<h2>Explanation:</h2><h3 /><h3>Oxygen- gains 2 electrons to form ions</h3><h3>Fluorine- gains 1 electron to form negative ions</h3><h3>Aluminum - loses three electrons to form ions</h3><h3>Calcium- loses 2 electrons in order to form ions</h3>

<h3>*Non metals gain electrons to form ions</h3><h3>*Metals loses electrons to form ions</h3>

6 0

3 years ago

Answer pleaseeeee for brainliest

BlackZzzverrR [31]

I would say the last maybe because my sister just got back from switerland not to long ago

8 0

4 years ago

What are epimers? How are glucose and manose epimers? Please Help!

katrin2010 [14]

Answer:

PLZ MARK ME BRIANLIEST I REALLY WANT IT PLZZZZZZ ISS GARIB KI DUA LAGEGI YARR PLZ

Explanation:

Glucose and galactose are monosaccharides that differ from one another only at position C-4. Thus, they are epimers that have an identical configuration in all the positions except in position C-4. ... Glucose and galactose are epimers that do not differ in position C-5 but differ in position C-4.

4 0

3 years ago

Read 2 more answers

Explain the regiochemistry of the Friedel-Crafts alkylation of 1,4-dimethoxybenzene. There are three positions available for the

defon

Answer:

In an electrophilic aromatic substitution (Friedel Crafts alkylation) first in the monoalkylation of the 1,4-dimethoxybenzenethe the methoxy groups redirects the substitution for ortho-para positions with respect to the electrophile that is going to enter (alkyl group) this is due to the increase in electron density in that position, that is , to the inductive effect.

According to the second incoming alkyl group there would be 3 positions available, from which it will choose the meta position in relation to the second methoxy group, since the alkyl group is a weak activator of the ortho meta positions and coincides with the position to which it redirects the second methoxy group.

7 0

3 years ago