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kkurt [141]
3 years ago
8

The boiling points of three substances are listed below. Which substance is most

Chemistry
1 answer:
Schach [20]3 years ago
7 0

Answer:

substance 1&substance 3

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Please help. I’m stuck
barxatty [35]
A cartilage doesn’t produce blood cells
4 0
3 years ago
And non-flammable gases<br> Noble gases are<br> that have low chemical
gtnhenbr [62]

Answer:

  • Noble gases are <u>odorless, colorless,</u> and nonflammable gases that have low chemical <u>reactivity</u><u>.</u>
  • The full <u>valence electron shells</u><u> </u>of these atoms make noble gases extremely <u>stable</u><u>.</u>
  • & they are <u>unlikely to form chemical bonds</u><u> </u>because they have little tendency to gain or lose

<u>electrons.</u>

5 0
3 years ago
What are water and carbon dioxide converted into by the end of
timama [110]

Answer:

sugar and oxygen

Explanation:

In photosynthesis, solar energy is harvested as chemical energy in a process that converts water and carbon dioxide to glucose. Oxygen is released as a byproduct.

6 0
3 years ago
Read 2 more answers
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
2. What ions are present in what ratio in a solution of aqueous calcium chloride?
Alenkasestr [34]

Answer:

\mathrm{Ca}^{2+} \text { and } \mathrm{Cl} \text { - ions are present in } 1: 2 \text { ratio in a solution of aqueous calcium chloride. }

Explanation:

Here in Calcium Chloride ionic bond is present in between calcium and chlorine atoms. As we know according to Octet rule calcium have two excess atoms and for matching nearest noble gas electronic configuration. It donate two electrons to gain more stability and form \mathrm{Ca}^{2+}, while chlorine is deficient from one electron to meet nearest noble gas electronic configuration therefore two chlorine atoms accept excess electron from calcium individually and form two\mathrm{Cl}^{-} ions.

\text { Equation is as follows: } \mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CaCl}_{2}

Hence aqueous solution of calcium chloride breaks the ionic bond pairing in one \mathrm{Ca}^{2+}and two\mathrm{Cl}^{-}ions: \mathrm{CaCl}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \quad \mathrm{Ca}^{2+}(\mathrm{ag})+2 \mathrm{Cl}(\mathrm{ag})

5 0
4 years ago
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