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tankabanditka [31]
4 years ago
5

An iron block of mass 18 kg is heated from 285 K to 318 K. If 267.3 kJ is required, what is the specific heat of iron? A. 450.00

B. 145.80 C. 158.77 D. 490.05
I am looking for an ABC or D not a long list of explanations
Chemistry
1 answer:
valkas [14]4 years ago
6 0

Answer:

  • <u>Option A. 450.00</u>

Explanation:

<u>1) Data:</u>

a) m = 18 kg

b) T₁ = 285 K

c) T₂ = 318 K

d) Q = 267.3 kJ

e) S = ?

<u>2) Principles and equations</u>

The specific heat of a substance is the amount of heat energy absorbed to increase the temperature of certain amount (gram, kg, or moles, depending on the definition or units) of the substance in 1 ° C or 1 K.

The mathematical relation between the specific heat and the heat energy absorbed is:

  • Q = m × S × ΔT

Where,

  • Q is the heat absorbed,
  • S is the specific heat, and
  • ΔT is the temperature increase (T₂ - T₁)

<u>3) Solution:</u>

<u>a) Substitute the data into the equation:</u>

  • 267.3 kJ = 18 kg × S × (318 K - 285 K)

<u>b) Solve for S and compute:</u>

  • S = 267.3 kJ / (18 kg × 33 K) = 0.45 kJ / (Kg . K)

The options have not units, but I notice that the first answer is 1,000 times the answer I obtained, so I will make a conversion of units.

<u>c) Convert to J /( kg . k):</u>

  • 0.45 kJ / (Kg . K) × 1,000 J / kJ = 450 J / (kg . K)

Now we can see that the option A is is the answer, assuming the units.

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Answer:

118 elements

Explanation:

Of these 118 elements, 94 occur naturally on Earth.

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3 years ago
If the mass ratio of nitrogen to hydrogen in ammonia is 4.7:1. if a sample of ammonia contains 10.0 gram of h, how many grams of
Tju [1.3M]

The amount, in grams, of N that the sample will contain will be 2.1 grams.

<h3>Stoichiometric mass ratio</h3>

According to the problem. the mass ratio of H and N in ammonia is 4.7:1.

In other words, every 4.7 grams of H in ammonia must have 1 gram of N.

Now, in a particular ammonia sample, 10 grams of H is present.

The amount of N present can be calculated from the standard mass ratio.

4.1 grams H = 1 gram N

10 grams H = x

4.1x = 1 x 10

        x = 10/4.1

             x = 2.1 grams

Thus, the mass of nitrogen in the ammonia sample will be 2.1 grams.

More on mass ratios can be found here: brainly.com/question/14577772

#SPJ1

6 0
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Which sound has waves with the greatest amplitude?'
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The pK1, pK2, and pKR of the amino acid lysine are 2.2, 9.1, and 10.5, respectively. The pK1, pK2, and pKR of the amino acid arg
almond37 [142]

Answer:

pH 9,8 is likely to work best for this separation

Explanation:

Ion exchange chromatography is a chemical process where molecules are separated by affinity to an ion exchange resin. To separate different aminoacids you must use the isoelectric point (That is the pH where the aminoacid will be in its neutral form).

For lysine, PI is:

pH = \frac{1}{2} (9,1+10,5) = 9,8

For arginine:

pH = \frac{1}{2} (9,0+12,5) = 10,75

At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.

Thus, <em>pH 9,8 is likely to work best for this separation</em>

<em></em>

I hope it helps!

5 0
3 years ago
Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres
____ [38]

Answer:

Initially 1.51\times 10^{-2} moles of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature =V_1=3.4 L

Number of moles in the container = n_1

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = V_2=5.11 L

Number of moles in the container after addition=n_2=2.28\times 10^{-2} mol

Applying Avogadro's law:

\frac{Volume}{Moles}=constant (at constant pressure and temperature)

\frac{V_1}{n_1}=\frac{V_2}{n_2}

n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}

n_1=1.51\times 10^{-2} mol

Initially 1.51\times 10^{-2} moles of nitrogen dioxide were in the container .

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