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jarptica [38.1K]
2 years ago
7

Lead is an important input in the production of crystal. If the price of lead decreases, then we would expect the supply of a. c

rystal to be unaffected. b. lead to increase. c. crystal to decrease. d. crystal to increase.
Chemistry
1 answer:
UkoKoshka [18]2 years ago
7 0

Answer: Option (b) is the correct answer.

Explanation:

Since, the input factor of production is lead and its price has come down. So, with decrease in price there will also occur a decrease in cost of production of the crystal.

And, when cost of production is low then it is possible to create more number of lead crystals due to which there will also occur an increase in supply of the crystal.

Therefore, we can conclude that lead is an important input in the production of crystal. If the price of lead decreases, then we would expect the supply of lead increases.

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Which of the following is a characteristic of a scientific theory
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Answer:

number one

Explanation:

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8 0
2 years ago
If 0.0714 moles of N2 gas occupies 1.25 L space, how many moles of N2 have a volume of 25.0 L? Assume temperature and pressure s
vfiekz [6]

Answer:

1.428 moles

Explanation:

If 0.0714 moles of N2 gas occupies 1.25 L space,

how many moles of N2 have a volume of 25.0 L?

Assume temperature and pressure stayed constant.

we experience it 0.0714 moles: 1.25L space

x moles : 25L of space

to get the x moles, cross multiply

(0.0714 x 25)/1.25

1.785/1.25 = 1.428 moles

8 0
3 years ago
In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

7 0
3 years ago
Solution (A) contains 0.01 M of NH4Cl; solution
ArbitrLikvidat [17]

Answer:

NH4Cl, NaCl, Ba(OH)2, NaOH

Explanation:

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NaCl is a neutral salt,formed by neutralization of a strong acid (HCl) with a strong base (NaOH). Hence, it will have a PH of 7.

Ba(OH)2 is a weak base. Therefore, it will have a PH between 8 and 10.

NaOh meanwhile, is a strong base. Therefore, it will have a PH between 10 to 13.

Hence, we have

NH4Cl < NaCl < Ba(OH)2 < NaOH

5 0
3 years ago
How much energy is required to raise the temperature of 0.2 kg of aluminum
Ann [662]
B. 538 j hoped this helps
3 0
3 years ago
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