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4 years ago

6

Where am I missing an electron?

1 answer:

shtirl [24]4 years ago

4 0

There should be three electrons.
Four neutrons and four protons.

let me know if i helped or if I was wrong

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In fall, leaves may change from green to yellow or red. Explain in your own words what is happening inside the leaf with regard

natima [27]

Answer:

The pigment that causes leaves to be green is chlorophyll. ... As chlorophyll goes away, other pigments start to show their colors. This is why leaves turn yellow or red in fall. In fall, plants break down and reabsorb chlorophyll, letting the colors of other pigments show through.

3 0

3 years ago

1. Write the structural formulas and give IUPAC names for all isorneric alcohols of molecular formula

EastWind [94]

Explanation:

Butanol (С4Н9OH)

Structural formula is :-

CH3-CH2-CH2-CH2-OH

6 0

3 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

4 years ago

at 55 years what mass remains of a 200.0 g sample of a radioactive isotope with a half life of 10.0 years

Harman [31]

Answer:

4.419 g

Explanation:

55 years is 5.5 half lives

200 g * (1/2)^5.5 = 4.419 g

3 0

1 year ago

Set the Room temp. to -10.0 °C and observe. What is happening?

Lady bird [3.3K]

Answer:

its getting colder

Explanation:

6 0

3 years ago

Read 2 more answers