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never [62]
3 years ago
11

A 200 g block of a substance requires 520 J of heat to raise its temperature from 25°C to 45°C.

Chemistry
1 answer:
harkovskaia [24]3 years ago
6 0

<h3>Answer: </h3>

The answer is gold.

<h3>Explanation:</h3>

Substance: Specific Heat (C) in joules per gram/degrees centigrade:

Water (ice) 2.05

Iron 0.46

Aluminium 0.90

Gold 0.13

Copper 0.39

Ammonia (liquid) 4.70

Ethanol 2.44

Gasoline 2.22

Water (liquid) 4.18

Water (vapor) 2.08

Air (25 degrees Celsius) 1.01

Oxygen 0.92

Hydrogen 14.30

To know that we clear the equation q=m*C*Δt for the specific heat (C) and solve the equation. Keep in mind that we need to use the heat (520 J) in Joules since the values from the question and the table are in Joules.

q/m∗ΔT = C = 520J/200g∗(45degC−25degC) =

0.13J/gdegC

As you can see, the substance that has this specific heat is GOLD.

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natali 33 [55]

Answer:

63.616

Explanation:

DATA

1. first atomic mass;m1=63

  1. second atomic mass;m2=65
  2. first percentage;p1= 69.2%
  3. second percentage me;p2=30.8%
  4. average mass;avg= ?

SOLUTION

avg=<u> (m1)(p1) + (m2)(</u><u>p2</u><u>)</u>

100

avg= <u>(63)(69.2) + (65)(30.8)</u>

100

avg= <u>4</u><u>3</u><u>5</u><u>9</u><u>.</u><u>6</u><u> </u><u>+</u><u> </u><u>2</u><u>0</u><u>0</u><u>2</u>

100

avg= <u>6361.6</u>

100

avg= 63.616

4 0
3 years ago
Read 2 more answers
The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.
Vikki [24]

Answer :  The solubility of this compound in g/L is 565.414\times 10^{-6}g/L.

Solution : Given,

K_{sp}=2.42\times 10^{-11}

Molar mass of MnCO_3 = 114.945g/mole

The balanced equilibrium reaction is,

                      MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3

At equilibrium                         s       s

The expression for solubility constant is,

K_{sp}=[Mn^{2+}][CO^{2-}_3]

Now put the given values in this expression, we get

2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L

The value of 's' is the molar concentration of manganese ion and carbonate ion.

Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.

s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L

Therefore, the solubility of this compound in g/L is 565.414\times 10^{-6}g/L.


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