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slava [35]
3 years ago
6

An atom contains three protons, two neutrons, and one electron. What is the charge of the nucleus?

Chemistry
1 answer:
andreev551 [17]3 years ago
7 0

Answer:

A.3+

Explanation:

️️️️️️️️️️️️️️

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At constant temperature the pressure on a 6.0 L sample of a gas is reduced from 2.0 atm to 1.0 atm. What is the new volume of th
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P1V1=P2V2, so P1V1/P2=V2.
2atm x 6.0 L/1.0 atm = 12.0 L
The new volume would be 12.0 Liters
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How many atoms are in 3.690 moles of oxygen
sweet [91]

1 mole of any substance contains Avogadro's number.  

So, 1 mole of O2= 6.023x10^23 molecules  

3 mole of O2= 6.023x10^23x3 molecules  

= 1.8069x10^24 molecules  

Each molecule of Oxygen has 2 atoms.  

therefore,  

1.8069x10^24 molecules= 1.8069x10^24 x 2 atoms  

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3 years ago
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Lao tzu believed that _____.
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The answer is b. propel should be guided by the way of nature
8 0
3 years ago
Calculate the amount of heat in kJ that is required to heat 25.0 g of ice from -25 °C to 105 °C in a closed vessel and sketch a
kolezko [41]

Answer:

The total amount of heat required for the process is 76.86 KJ

Explanation:

We can divide the process in 5 parts, in which we can calcule each amount of heat required (see attached Heating curve):

(1) Ice is heated from -25ºC to 0ºC. We can calculate the heat of this part of the process as follows. Note that we must convert J in KJ (1 KJ= 1000 J).

Heat (1) = mass ice x Specific heat ice x (Final temperature - Initial Temperature)

Heat (1) =25 g x 2.11 J/g.ºC x \frac{1 KJ}{1000 J} x (0ºC-(-25º)

Heat (1) = 1.32 KJ

(2) Ice melts at ºC (it becomes liquid water). This is heating at constant temperature (ºC), so we use the melting enthalphy (ΔHmelt) and we must use the molecular weight of water (1 mol H₂O = 18 g):

Heat (2) = mass ice x ΔHmelt

Heat (2)= 25 g  x  \frac{6.01KJ} {1 mol H2O} x \frac{1 mol H2O}{18 g}

Heat (2)= 8.35 KJ

(3) Liquid water is heated from 0ºC to 100 ºC:

Heat (3)= mass liquid water x Specific heat water x (Final T - Initial T)

Heat (3)= 25 g x 4.18 J/gºC x 1 KJ/1000 J x (100ºC - 0ºC)

Heat (3)= 10.45 KJ

(4) Liquid water evaporates at 100ºC (it becomes water vapor). This is a process at constant temperature (100ºC), and we use boiling enthalpy:

Heat (4)= mass water x ΔH boiling

Heat (4)= 25 g x \frac{40.67 KJ}{mol H20} x \frac{1 mol H20}{18 g}

Heat (4)= 56.49 KJ

(5) Water vapor is heated from 100ºC to 105ºC. We use the specific capacity of water vapor:

Heat (5)= mass water vapor x Specific capacity vapor x (Final T - Initial T)

Heat (5)= 25 g x 2.00 J/g ºC x 1 KJ/1000 J x (105ºC - 100ºC)

Heat (5)= 0.25 KJ

Finally, we calculate the total heat involved in the overall process:

Total heat= Heat(1) + (Heat(2) + Heat(3) + Heat(4) + Heat(5)

Total heat= 1.32 KJ + 8.35 KJ + 10.45 KJ + 56.49 KJ + 0.25 KJ

Total heat= 76.86 KJ

3 0
4 years ago
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