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2 years ago

For an atom (which is neutral), which of these must be

Chemistry

1 answer:

jeyben [28]2 years ago

8 0

Answer:

The number of protons is equal to the number of electrons

Explanation:

For an atom to be neutral, all the charges have to equal zero. Because protons are charged +1 and electrons are charged -1, the number of protons has to be equal to the number of electrons for the charges to cancel out.

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A student assisting with the experiment would observe all of the following about the electron transport chain EXCEPT:A. Electron

zheka24 [161]

Answer:

C. All electron carriers are mobile and hydrophobic

Explanation:

Hello,

In this case, it is widely known that the electron carriers move inside the inner mitochondrial membrane and consequently move electrons from one to another. In such a way, they are mobile, therefore they are largely hydrophobic as long as they are inside the membrane.

For instance, the cytochrome c is a water-soluble protein in a large range, therefore, the answer is: C. All electron carriers are mobile and hydrophobic.

Best regards.

5 0

2 years ago

PLZ HELP Organisms on earth are ________ based life forms. - oxygen - hydrogen - carbon

Bess [88]

Answer:

Carbon based life forms

5 0

2 years ago

Read 2 more answers

What is the ph of a solution that is 0.50 m in propanoic acid and 0.40 m in sodium propanoate. (ka for propanoic acid = 1.3 x 10

max2010maxim [7]

The pH is given by the Henderson - Hasselbalch equation:
 pH = pKa + log([A-]/[HA])
 pH = -log(1.3 x 10^-5) + log(0.50/0.40)
 pH = 4.98
The answer to this question is 4.98.

3 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

What is the density of object A? Does it sink or float in water?

Oksana_A [137]

Answer:

~1.5 g/cm3 and it does NOT float in water.

Explanation:

If you look at the graph, Object A weighs ~6 grams and is ~4 cm3 in volume

Density = Mass/Volume

So 6 grams/4 cm3 = 1.5 g/cm3

Water has a density of 1 g/cm3 and because Object A density is higher than that of water, it sinks.

3 0

2 years ago