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3 years ago

15

What is the total charge of the nucleus of a nitrogen atom

2 answers:

deff fn [24]3 years ago

8 0

This easy man.its positive charge. i learned this at school.

Soloha48 [4]3 years ago

4 0

+ positive charge
because nucleus always has a positive charge due to proton

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explain using diagrams how potassium forms the compound potassium flouride when it reacts with flourine

Katarina [22]

Answer:

The answer to your question is given below.

Explanation:

Potassium (K) has 19 electrons with electronic configuration of 2, 8, 8, 1.

Fluorine (F) has 9 electrons with electronic configuration of 2, 7.

Fluorine needs 1 electron to complete it's octet configuration.

Hence, potassium (K), will lose 1 electron to fluorine (F) to form potassium ion (K+) with electronic configuration of 2, 8, 8. The fluorine atom (F) will receive the 1 electron from potassium to form the fluoride ion (F-) with electronic configuration of 2, 8.

**** Please see attached photo for further details.

7 0

2 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

3 years ago

Given the following balanced equation, determine the rate of reaction with respect to [SO2].

Gnoma [55]

Answer:

Rate = -1/2 Δ[SO<sub>2</sub>]/Δt

so its gonna be (in more simple terms) rate= -1/2Δ(SO2)/Δt

Explanation:

7 0

2 years ago

Be sure to answer all parts.

Marianna [84]

a)3^2 = 9 times

b)2^2*2 = 8 times

c)(1/2) = half times

7 0

2 years ago

What did Mendel do that made his predictions so accurate

Hoochie [10]

He worked with large numbers of plants.

8 0

3 years ago

Read 2 more answers