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wel
3 years ago
11

2.5 piece of lithium is dropped into a 500 g sample of water at 22.1 C.ten temperature of the water increases to 23.5 C. How muc

h heat energy is released in the reaction
Chemistry
1 answer:
Dominik [7]3 years ago
3 0

Answer:

2.6 kJ  

Explanation:

The formula for the amount of heat (q) absorbed by the water is

q = mCΔT

1. Calculate ΔT

ΔT = 23.5 °C - 22.1 °C = 1.4 °C

2. Calculate q

q₂ = mCΔT = 500 g × 4.184 J·°C⁻¹g⁻¹ × 1.4 °C = 2900 J = 2.9 kJ

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Melting 

Melting is a change in property of matter from solid to liquid

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5 0
3 years ago
A student was asked to determine the densities of the four metals listed. The student was asked to determine the density of an u
Iteru [2.4K]

Answer:

D is correct

Explanation:

because

we know that

density of lead is 11.36 g/cm3

and

density of tin is 7.31 g/cm3

so..

density of alloy by mixing 50/50

=(11.36+7.31)/2 g/cm3

=18.67/2 g/cm3

=9.33 g/cm3

5 0
3 years ago
On headedama be prepared​
mixer [17]

Answer:

yes

Explanation:

yea

3 0
2 years ago
When can a compound be treated like an element when balancing chemical equations?
Marrrta [24]
A compound can be treated like an element when balancing chemical equations when : A. when there is no change to the atoms in the compound

If they remain together during the reaction, you can treat a compound as a singe atom.

hope this helps
7 0
3 years ago
Read 2 more answers
RATE LAW QUESTION !
vivado [14]
In general, we have this rate law express.:

\mathrm{Rate} = k \cdot [A]^x [B]^y
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4)  by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation \Delta \mathrm{Rate} = \Delta [B]^y

3.09 = (3)^y \implies y \approx 1

To this point, \mathrm{Rate} = k \cdot [A]^x [B]^1

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for \Delta \mathrm{Rate} = \Delta [A]^x

4=  (2)^x \implies x = 2

the rate law is

Rate = k·[A]²[B]
6 0
3 years ago
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