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Reptile [31]
3 years ago
11

A phase diagram assumes ______ is kept constant

Chemistry
2 answers:
bogdanovich [222]3 years ago
8 0
The answer is b i think

asambeis [7]3 years ago
8 0
Pressure 





Hope that helps <3
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How many atoms of hydrogen are found in the compound, NH4C2H302?
Mariulka [41]
<h2>Answer:</h2>

7 hydrogen atoms

<h2>Explanation:</h2>

N<em><u>H4</u></em>C2<em><u>H3</u></em>02

In this problem we see the hydrogen atom twice, along with the numbers 4 and 3 next to them. (as shown above in bold & underlined)

So, in order to find how many there are in all you add both hydrogen atoms together-

H4+H3= H7

therefore, there are 7 hydrogen atoms in all

8 0
3 years ago
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
Eutrophication results from an excess of _____.
gladu [14]
Eutrophication results from an excess of 

a. phosphates

3 0
3 years ago
Read 2 more answers
How do you know when a solution is saturated? _____________________________________________________ When a solution is saturated
mote1985 [20]

Answer:

If more solute is added and it does not dissolve, then the original solution was saturated

If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturate

Explanation:

5 0
3 years ago
Calculate the density a rectangular block of a metal whose length is 8.335cm width is 1.02cm height is 0.982cm and mass is 62.35
Ostrovityanka [42]

Answer:

Density rectangular block = 7.47 (Approx) gm/cm³

Explanation:

Given:

Length = 8.335 cm

Width = 1.02 cm

Height = 0.982 cm

Mass = 62.3538 gm

Find:

Density rectangular block

Computation:

Volume of block = lbh

Volume of block = (8.335)(1.02)(0.982)

Volume of block = 8.3486 cm³

Density = Mass / Volume

Density rectangular block = 62.3538 / 8.3486

Density rectangular block = 7.47 (Approx) gm/cm³

8 0
3 years ago
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