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tatiyna
3 years ago
13

Which substance is a gas at 20˚C and one atmosphere of pressure? A. C B. O3 C. Ca D. I2

Chemistry
1 answer:
jasenka [17]3 years ago
3 0

Answer:

O3

Explanation:

The answer to the question is O3 .

Hope this helped! Have an amazing day :)

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How many Mg atoms are present in 170g of Mg
PIT_PIT [208]
In order to calculate the number of atoms, we must first know the number of moles present. And
moles = (mass present) / (molecular mass)
Therefore, the moles of Mg present are
170 / 24 = 7.08
The number of atoms in a mole of substance is given by Avagadro's Number which is 6.02 x 10^23
Since there are 7.08 moles, there are:
7.08 * 6.02*10^23
= 4.26 * 10^24 atoms
8 0
3 years ago
If the solubility of sugar at 25ºC is 211 g/100 g H2O, what mass of sugar can be dissolved in 300 g of H2O?
ladessa [460]

Answer:

633 grams of sugar can be dissolved in 300 g of H₂O

Explanation:

Solubility is the measure of the ability of a certain substance to dissolve in another and form a homogeneous system.  Solubility is then the maximum amount of a solute that a solvent can receive and is expressed by concentration units.

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

Then:

x=\frac{c*b}{a}

You can apply the rule of three as follows: if by definition of solubility in 100 grams of H₂O there are 211 grams of sugar, in 300 g of H₂O how much sugar is there?

sugar=\frac{300 grams of H_{2}O *211 grams of sugar}{100 grams of H_{2}O}

sugar= 633 grams

<u><em>633 grams of sugar can be dissolved in 300 g of H₂O</em></u>

7 0
3 years ago
Read 2 more answers
The electronic configuration of an element is given below.
Shalnov [3]

Answer:

It is reactive because it has to gain an electron to have a full outermost energy level.

Explanation:

The electron configuration of oxygen is 1s2,2s2 2p4.

Oxygen is in group six in the periodic table so it has six electrons in its valence shell. This means that it needs to gain two electrons to obey the octet rule and have a full outer shell of electrons (eight).  

3 0
2 years ago
Calculate the oxidation numbers of the elements in this equation<br> 2 Al2O3 → 4 Al + 3 O2
riadik2000 [5.3K]

Explanation:

Answer

Open in answr app

The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.

Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O )  = 0

2× ( Oxidation number of Al) +3(−2)=0

2× ( oxidation number of Al) +6

∴ Oxidation number of Al =+3

5 0
3 years ago
What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances
belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

4 0
3 years ago
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