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xeze [42]
3 years ago
15

The temperature of a system rises by 45°C during a heating process. Express this rise in temperature in Kelvin. (Round the final

answer to the nearest whole number.)
Chemistry
1 answer:
postnew [5]3 years ago
4 0

Answer:

45K

Explanation:

Rise in temperature = Final - initial temperature.

temperature in K = Temperature in Celsius + 273

for Celsius; T2 -T1 =45°C

for kelvin; T2+273 -(T1+273) = ?

                T2+273 -T1-273 =?

                T2-T1 = ?

               T2-T1 =45k

hence ΔT(K) = ΔT(°C) (temperature difference in Celsius is equal to temperature difference in kelvin)

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3 0
2 years ago
2c+02=2CO2. The moles of co2 produced when 0.25 moles of O2 react is?​
Sophie [7]
<h3>Answer:</h3>

\displaystyle 0.5 \ mol \ CO_2

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Moles
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<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2C + O₂ → 2CO₂

[Given] 0.25 moles O₂

[Solve] moles CO₂

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol O₂ → 2 mol CO₂

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up:                                                                                                     \displaystyle 0.25 \ moles \ O_2(\frac{2 \ mol \ CO_2}{1 \ mol \ O_2})
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7 0
3 years ago
If an ice cube weighing 25.0 g with an initial
riadik2000 [5.3K]

Answer:

11

∘

C

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at

0

∘

C

to liquid at

0

∘

C

.

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

q

1

+

q

2

=

−

q

3

(

1

)

, where

q

1

- the heat absorbed by the solid at

0

∘

C

q

2

- the heat absorbed by the liquid at

0

∘

C

q

3

- the heat lost by the warmer water sample

The two equations that you will use are

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of water, equal to

4.18

J

g

∘

C

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

and

q

=

n

⋅

Δ

H

fus

, where

q

- heat absorbed

n

- the number of moles of water

Δ

H

fus

- the molar heat of fusion of water, equal to

6.01 kJ/mol

Use water's molar mass to find how many moles of water you have in the

100.0-g

sample

100.0

g

⋅

1 mole H

2

O

18.015

g

=

5.551 moles H

2

O

So, how much heat is needed to allow the sample to go from solid at

0

∘

C

to liquid at

0

∘

C

?

q

1

=

5.551

moles

⋅

6.01

kJ

mole

=

33.36 kJ

This means that equation

(

1

)

becomes

33.36 kJ

+

q

2

=

−

q

3

The minus sign for

q

3

is used because heat lost carries a negative sign.

So, if

T

f

is the final temperature of the water, you can say that

33.36 kJ

+

m

sample

⋅

c

⋅

Δ

T

sample

=

−

m

water

⋅

c

⋅

Δ

T

water

More specifically, you have

33.36 kJ

+

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

0

)

∘

C

=

−

650

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

25

)

∘

C

33.36 kJ

+

418 J

⋅

(

T

f

−

0

)

=

−

2717 J

⋅

(

T

f

−

25

)

Convert the joules to kilojoules to get

33.36

kJ

+

0.418

kJ

⋅

T

f

=

−

2.717

kJ

⋅

(

T

f

−

25

)

This is equivalent to

0.418

⋅

T

f

+

2.717

⋅

T

f

=

67.925

−

33.36

T

f

=

34.565

0.418

+

2.717

=

11.026

∘

C

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

T

f

=

11

∘

C

Explanation:

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125 Each half life it divides by 2 the amount
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