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ira [324]
3 years ago
10

Ionic compounds are solid at room temperature because they have

Chemistry
1 answer:
Bumek [7]3 years ago
5 0

Answer:option A High melting points

Explanation:

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True. Nuclear fusion of hydrogen to form helium occurs naturallyin the sun and other stars. It takes place only at extremely high temperatures.
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3 years ago
It takes 330 joules of energy to raise the temperature of 24.6 gbenzene from 21 degrees Celsius to 28.7 degrees Celsius at const
lilavasa [31]

Given :

Energy , E = 330 J .

Initial temperature , T_i=21^oC .

Final temperature , T_f=24.6^oC .

Mass of benzene , m = 24.6 g .

To Find :

The molar hear capacity of benzene at constant pressure .

Solution :

Molecular mass of benzene , M = 78 g/mol .

Number of moles of benzene :

n=\dfrac{24.6}{78} \ mol\\\\n=0.32 \ mol

Energy required is given by :

q=nC_p\Delta T\\\\330=0.32\times C_p\times (28.7-21)\\\\C_p=\dfrac{330}{0.32\times 7.7}\ J\ mol^{-1}^oC^{-1} \\\\C_p=133.9\ J\ mol^{-1}^oC^{-1}

Hence , this is the required solution .

8 0
3 years ago
Milk of magnesia helps to neutralize stomach acid because it is a
dimulka [17.4K]
*base
a base is added to an acid to neutralize it.
3 0
3 years ago
Read 2 more answers
This table resembles a condensed version of the modern periodic table. Using the full periodic table as a reference, type the sy
Mekhanik [1.2K]

re write the question.

Explanation:

it seems incomplete.

5 0
2 years ago
1. A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C. If the
Maurinko [17]

Answer:

Mass of aluminium in sample = 3.591 g ≅ 3.6 grams

Explanation:

Given that,  A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C.

the specific heat of aluminum is 0.900 J/g- °C

The relation between heat absorbed and change in temperature is given by,   Q = msΔT.

where Q = heat absorbed

            m = mass of the substance

            s = specific heat of substance

          ΔT  = change in temperature

Now, in our case, Q = 50.1 J ; s = 0.900 J/g- °C; ΔT= 35.5-20 = 15.5°C

⇒ m =  \frac{Q}{s(T_{2} -T_{1}) }

⇒ m = \frac{50.1}{0.900(15.5)} = 3.591 g ≅ 3.6 g

⇒ m ≅ 3.6 g

5 0
3 years ago
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