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Scorpion4ik [409]
2 years ago
6

How many grams are in 9.05 x 1023 atoms of silicon?

Chemistry
1 answer:
Sergeu [11.5K]2 years ago
6 0

Explanation:

Number of moles(n)=Number of atoms(N)/Avogadro's constant.

Avogadro's constant=6.02×10²³

so we have

n=9.05×10²³/6.02×10²³

n=1.0503moles.

n=mass/molar mass

1.0503=mass/28

mass=1.0503×28

mass=29.4084g

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How does a battery become depleted or out of charge? * -the number of protons and electrons at both terminals are not equal -the
kompoz [17]

Answer:

the number of protons and neutrons at both terminals are equal

Explanation:

When the number of positive charge and negative charge are both equally the terminal, it becomes neutral and out of charge, because first it undergo enough chemical reaction and there is no remaining tendency for positive and negative charges to get separated. When this tendency dies, the battery also will run out of charge.

3 0
3 years ago
What is the mass of a 300 ml sample of gaseous hydrogen chloride at 2.0 atm and 30c?
kap26 [50]
Use PV = nRT
(2 atm)(.3 liters) = n(8.314 mol*K)(303°K)

.6 = n(2519.142)

Divide by 2519.142

n = .00023818 mols of HCl * 36.46g of HCl/ 1 mol of HCl

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3 0
3 years ago
Help?......please???
Flura [38]

Answer:im in middle school in know this but w x y = d

Explanation:so your answer is d :)

5 0
3 years ago
Example of chemical change?
musickatia [10]

Answer:

Burning of paper and log of wood.

Digestion of food.

Boiling an egg.

Chemical battery usage.

Electroplating a metal.

Baking a cake.

Milk going sour.

Various metabolic reactions that take place in the cells.

hope this helps!

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3 0
3 years ago
The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia
Savatey [412]

Answer:

\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0
3 years ago
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