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mihalych1998 [28]
3 years ago
9

What information does the first quantum number of an electron give?

Chemistry
1 answer:
bazaltina [42]3 years ago
4 0

Answer:

с

Explanation:

the first quantum number of an electron gives the information about the energy level the electron is in

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On a periodic table, element names are represented by:
alina1380 [7]

Answer: Symbols

Explanation:

4 0
3 years ago
If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol
igomit [66]

Answer:

\Delta _RH=4x10^{-3}\frac{kJ}{mol}

Explanation:

Hello,

In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}

Best regards.

5 0
3 years ago
I need help please j
Sidana [21]
Reactants Hydrogen: 5
Products Hydrogen: 5

Reactants Carbon: 3
Products Carbon: 3

Reactants Oxygen: 4
Products Oxygen: 5
3 0
3 years ago
Will Maynez burns a 0.6-g peanut beneath 50 g of water, which increases in temperature from 22°C to 50°C. (The specific heat cap
Neko [114]

Answer:

40% of  the energy release by the peanut is 3500 calories

Explanation:

One calorie is defined as the amount of energy required to increase the temperature of one gram of water for one degree Celsius (or one Kelvin)

Equation for energy gain by water is

Q = mcΔT

where, m is the mass of the object

c is the specific heat capacity

ΔT is the change in temperature

c =  1.0 cal/g?°C.

m = 50 g

ΔT = 50°C - 22°C

    = 28°C

Q = (50)× (1)× (28)

  = 1400calories

The peanut contain 1400calories of energy .

amount that 40% of energy is released to water ,

so,

Q = 1400 calories / 0.4

= 3500 calories

Therefore, 40% of  the energy release by the peanut is 3500 calories

7 0
3 years ago
How much heat (in kj) is required to warm 13.0 g of ice, initially at -12.0 ∘c, to steam at 113.0 ∘c? the heat capacity of ice i
olga nikolaevna [1]
The total amount of heat required is the sum of all the sensible heat and latent heats involved in bringing the ice to a desired temperature and state. The latent heat of fusion and vaporization of water 333.55 J/g and 2260 J/g, respectively. Solving for the total amount of heat,
                   total amount of heat = 13.0 g (2.09 J/gC)(12) + 13(333.55 J/g) + 13.0 g (4.18 J/gC)(100 - 0) + (13.0 g)(2260 J/g) + (13 g)(2.01 J/g)(113-100) 
                                        = 39815.88 J 
                                        = 39.82 kJ
5 0
3 years ago
Read 2 more answers
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