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Nimfa-mama [501]
3 years ago
12

an electric heater having resistance equal to 5 ohm is connected to electric source . If it produces 180 J of heat in one second

,find the potential difference across the electric heaterl​
Chemistry
1 answer:
Sindrei [870]3 years ago
3 0

Answer:

V = 30 volts

Explanation:

Given that,

The resistance of an electric heater, R = 5 ohm

Heat produced is 180 J in one second

We need to find the potential difference of the electric heater. The heat produced is given by :

E=I^2Rt\\\\or\\\\E=\dfrac{V^2}{R}t\\\\V=\sqrt{\dfrac{ER}{t}} \\\\V=\sqrt{\dfrac{180\times 5}{1}} \\\\V=30\ V

So, the potential difference across the heater is 30 volts.

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A 26.08 g mixture of zinc and sodium is reacted with a stoichiometric amount of sulfuric acid. The reaction mixture is then reac
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Answer:

Molar percent of sodium in original mixture is 88,50%

Explanation:

The last reaction is:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl

The moles of BaCl₂ are:

0,132L × 3,80M = 0,502 moles of BaCl₂

As the amount of BaCl₂ is the maximum possible to produce BaSO₄, the moles of BaCl₂ must be the same than moles of Na₂SO₄.

0,502 moles of BaCl₂ ≡ 0,502 moles of Na₂SO₄

These moles of Na₂SO₄ comes from:

2 Na + H₂SO₄ → Na₂SO₄ + H₂

As the reaction is in stoichiometric amounts, moles of Na are twice the moles of Na₂SO₄

0,502 moles of Na₂SO₄ ×\frac{2molesNa}{1moleNa_{2}SO_{4}}× 22,99 g/mole = 23,08 g of Na

Molar percent of sodium in original mixture is:

\frac{23,08g}{26,08g}*100 = <em>88,50% </em>

I hope it helps

4 0
3 years ago
How can you tell if a chemical is a metallic bond?
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Explanation:

The given data is as follows.

      Mass = 27.9 g/mol

As we know that according to Avogadro's number there are 6.023 \times 10^{26} atom present in 1 mole. Therefore, weight of 1 atom will be as follows.

            1 atoms weight = \frac{38}{6.023 \times 10^{26}}    

In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.

Therefore, total weight of atoms in a unit cell will be as follows.

            = \frac{8 \times 27.9 g/mol}{6.023 \times 10^{26}}

            = 37.06 \times 10^{-26}

Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.

                   = a^{3}

                   = (0.503 \times 10^{-9})^{3}

                   = 0.127 \times 10^{-27} m^{3}

Formula to calculate density of diamond cell is as follows.

               Density = \frac{mass}{volume}

                             = \frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}

                            = 2918.1 g/m^{3}

or,                         = 0.0029 g/cc       (as 1 m^{3} = 10^{6} cm^{3})

Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.

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3 years ago
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