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Basile [38]
3 years ago
10

A 50.0 g sample of an unknown substance, initially at 20.2 °C, was heated with 1.55 kJ of energy. The final temperature of the s

ubstance was 125.0 °C. Determine the specific heat of this substance.
Chemistry
1 answer:
valentina_108 [34]3 years ago
7 0

Answer:

0.296j/g⁰c

Explanation:

we have the following information from this question before us.

mass iv substance = 50grams

we have initial temperature ti = 20.2⁰c

final temperature = 125⁰c

the energy that was provided = 155kj

we proceed with this formula

energy = mcΔT

1.55x10³ = 50 x c x (125-20.2)

1.55x10³ = c x 50gm x 104.8k

we divide through to get c

c = 1.55x10³/50g x 104.8

c = 0.296J/g⁰c

that is the specific heat of this substance.

thank you!

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poizon [28]

Answer:-

(a) 3.5

(b) 3

Explanation:-

2KClO3 --> 2KCl + 3 O2

From the equation we see that 2 moles of KClO3 gives 2 moles of KCl.

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Again

3 moles of O2 are produced with 2 moles of KCl.

If 4.5 moles of O2 produced then

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How many dimes could you trade for 360 pesos? $1 = 1500 pesos.
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Answer:

The correct option is (d).

Explanation:

It is given that,

1$ = 1500 pesos

We need to convert 360 pesos into dimes

We can convert 360 pesos to dollars as follows:

360\ \text{pesos}=\$\dfrac{1}{1500}\times 360\\\\=$0.24

360 pesos is equal to $0.24

Also, 1 dollar = 10 dimes

We can covert 0.24 dollar to dimes as follows :

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0.24 dollar = 2.4 dimes

or

360 pesos = 2.4 dimes

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3 years ago
a sample of water with a mass of 648.00 kg at 298 K is heated with 87 kh of energy. the specific heat of water is 1 J-1 kg K-1.
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q = mCΔT

The correct specific heat capacity of water is <em>4.187 kJ/(kg.K)</em>.

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Tf = Ti + ΔT = 298 K + 0.032 K = 298.032 K


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