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3 years ago

Which is a gas at room temperature A.aluminum B.nitrogen C.potassium D.sodium

Chemistry

1 answer:

3241004551 [841]3 years ago

5 0

Answer:

B.nitrogen

Explanation:

Nitrogen is a gas at room temperature.

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The radius of an indium atom is 0.163 nm. If indium crystallizes in a face-centered unit cell, what is the length of an edge of

adoni [48]

<u>Answer:</u> The edge length of the unit cell is 0.461 nm

<u>Explanation:</u>

We are given:

Atomic radius of iridium = 0.163 nm

To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:

$a=2\sqrt{2}R$

Putting values in above equation, we get:

$a=2\sqrt{2}\times 0.163=0.461nm$

Hence, the edge length of the unit cell is 0.461 nm

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3 years ago

What is the reasoning behind Forgacs’s ideas onbiofabrication?

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3 years ago

Give the name and symbol of the prefixes used with SI units to indicate multiplication by the following exact quantites.

Bingel [31]

Answer:

A. Kilo , K, multiplication by 1000

B. Centi, c

C. Deci, d

D. Mili, m

E. Mega, M

F. Micro, u

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3 years ago

Did democritus belive atoms retain there identity as a chemical reaction

mihalych1998 [28]

No He believed tiny particles were invisible and couldn't be changed....So No The person that believed in this was Dalton .

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3 years ago

Read 2 more answers

The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:

Nimfa-mama [501]

Answer:

Enthalpy change for the reaction is -67716 J/mol.

Explanation:

Number of moles of $AgNO_{3}$ in 50.0 mL of 0.100 M of $AgNO_{3}$

= Number of moles of HCl in 50.0 mL of 0.100 M of HCl

= $\frac{0.100}{1000}\times 50.0$ moles

= 0.00500 moles

According to balanced equation, 1 mol of $AgNO_{3}$ reacts with 1 mol of HCl to form 1 mol of AgCl.

So, 0.00500 moles of $AgNO_{3}$ react with 0.00500 moles of HCl to form 0.00500 moles of AgCl

Total volume of solution = (50.0+50.0) mL = 100.0 mL

So, mass of solution = ( $100.0\times 1.00$ ) g = 100 g

Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)

= $\frac{-m_{solution}\times C_{solution}\times \Delta T_{solution}}{0.00500mol}$

= $\frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}$

= -67716 J/mol

[m = mass, c = specific heat capacity, $\Delta T$ = change in temperature and negative sign is included as it is an exothermic reaction]

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3 years ago