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Science Question Answer (5 Marks)

Atoms and Molecules · Maharashtra Board STD 9 (MSBSHSE - English Medium). 21 questions.

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Question Answer (5 Marks)

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks
State the law of constant proportions. Give one example to illustrate this law.
Answer
Law of constant proportion given by PROUST states that "A chemical compound always consists of the same elements combined together in the same proportion by mass."
For example: If we decompose 100 gms of pure water by passing electricity through it, then 11gms of hydrogen and 89gms of oxygen are obtained. Now, if we repeat this experiment by taking pure water from different sources (like river, sea, well, etc.), the same masses of hydrogen and oxygen elements are obtained in each case. They are always combined together in the same constant proportion of 11 : 89 or 1 : 8 by mass. And this is the law of constant proportions.
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Question 25 Marks
What mass of nitrogen $N_2$_, will contain the same number of molecules as $1.8g$ of water, $H_2O$?
(Atomic masses: $N = 1\ 4u; H = 1\ u; O = 16\ u$)
Answer
Mass of water $=1.8 g$
We know that equal moles of all substance contain equal number of molecules ( $1$ mole of all substances contains $6.022 \times 10^{23}$ molecules).
The first step will be to convert $1.8\ g$ water into moles.
$1 mole \text { of water }=(2 \times 1)+(1 \times 16)=18 g$
$18 g \text { water }=1 \text { mole }$
$1.8 g \text { water }=0.1 \text { moles }$
Since, equal moles of all substances contain equal number of molecules, $0.1$ moles of water will have the same number of molecules as present in $0.1$ moles of nitrogen.
The next step will be to find the mass of $0.1$ moles of nitrogen in grams.
$1$ mole of nitrogen $=28\ g$
$0.1$ moles of nitrogen $=28 \times 0.1=2.8 g$
Hence, $2.8\ g$ nitrogen will contain the same number of molecules as present in $1.8\ g$ water.
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Question 35 Marks
In an experiment, $4.90\ g$ of copper oxide was obtained form $3.92\ g$ of copper. In another experiment, $4.55\ g$ of copper oxide gave, on reduction, $3.64\ g$ of copper. Show with the help of calculations that these figures verify the law of constant proportions.
Answer
Rection 1:$\ \ 2\text{Cu} \ \ \ \ \ \ +\ \ \ \ \ \ \ \ \ \ \ \text{O}_2\ \ \ \ \ \ \ \ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ }\ \ \ \ 2\text{CuO}\\ ^{(3.92\text{gm})} \ \ \ \ \ \ \ \ ^{(4.90-3.92\ =\ 0.98\text{gm})}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{(4.90\text{gm})}$
$\frac{3.92}{3.92}=1,\ \frac{0.98}{3.92}=0.25,\ \frac{4.90}{3.92}=1.25$
So, $1$ equivalent of Cu reacts with $0.25$ equivalent of $O_2$ to from $1.25$ equivalent of copper oxide.
Reaction 2:
$2\text{CuO}\ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ }\ \ \ \ 2\text{Cu}\ \ \ +\ \ \ \ \ \ \ \ \ \ \text{O}_2\\ ^{(4.55\text{gm})}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{(3.64\text{gm})}\ \ \ \ \ \ ^{(4.55-3.64\ =\ 0.91\text{gm})}$
$\frac{4.55}{3.64}=1.25,\ \frac{3.64}{3.64}=1,\ \frac{0.91}{3.64}=0.25$
Here again, one can see that $1.25$ equivalent of $CuO$ decomposed to form $1$ equivalent of $Cu$ and $0.25$ equivalent of oxygen.
Hence, law of constant proportion is verified.
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MCQ 45 Marks
Write the cations and anions present, if any, in the following:
a.$CH_3COONa$
b.$NaCl$
c.$H_2$
d.$NH_4NO_3$
  • A

     
  • B

     
  • C

     
Answer
  1. $CH_3COONa$: $Na^+$(cation) and $CH^3COO^-$(anion)
  2. $NaCl$: $Na^+$(cation) and $Cl^-$(anion)
  3. $H_2$: It is a covalent molecule. So, cation and anion are not present.
  4. $NH_4NO_3$ : $NH_{4+}$(cation) and $NO_{3-}$(anion)
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Question 55 Marks
Which has more atoms, $50\ g$ of aluminium or $50\ g$ of iron? Illustrate your answer with the help of calculations.
(Atomic masses: $Al =27 u ; Fe =56 u$ )
Answer
$1$ mole of aluminium weighing $27\ g$ has $=6.022 \times 10^{23}$ atoms of Al
So, $1\ g$ of Al has $=0.22 \times 10^{23}$ atoms of Al
Hence, $50\ g$ of Al will have $=50 \times 0.22 \times 10^{23}$ atoms of Al $=11 \times 10^{23}$ atoms of Al
$1$ mole of iron weighing 56 g has $=6.022 \times 10^{23}$ atoms of Fe
So, 1 g of Fe has $=0.10 \times 1023$ atoms of Fe
Hence, $50\ g$ of Fe will have $=50 \times 0.10 \times 10^{23}$ atoms of $Fe =5 \times 10^{23}$ atoms of Fe
Thus, $50\ g$ of AI has more no. of atoms as compared to $50\ g$ of Fe .
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Question 65 Marks
If one gram of sulphur contains $x$ atoms, calculate the number of atoms in one gram of oxygen element.
(Atomic masses: $S = 32\ u; O = 16\ u)$
Answer
Given,
$1\ g$ sulphur contains $x$ atoms
We know that equal moles of all substances contain equal number of molecules ($1$ mole of all substances contains $6.022 \times 10^{23}$ molecules).
Therefore, the ratio of molecules in sulphur and oxygen will be the same as their ratio of moles.
Therefore, 1g of $\text{S} = \frac{1}{32}$ moles of S Now, $\frac{1}{32}$ moles of $S$ contain $x$ molecules (given).
Since equal number of molecules are present in equal moles of all substances,
$\frac{1}{32}$ moles of oxygen will also contain $x$ number of molecules.
Number of moles in $1g\ $ of oxygen:
$1 mole of O = 16\ g$
$16\ g O = 1 mole of O$
Therefore, $1\ g$ $\text{O}_2=\frac{1}{16}$ moles of $O_2$​​​​​​​
We have already stated that $\frac{1}{32}$ moles of oxygen contain $x$ molecules;
therefore,$\frac{1}{16}$ mole of oxygen will contain $=\frac{\text{x}\ \times\ 32}{16} = 2 \text{x molecules}$
Hence, $1$ gram oxygen consists of $2x$ the number of sulphur molecules.
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Question 75 Marks
An element X forms the following compounds with hydrogen, carbon and oxygen:
$H_2X, CX_2, XO_2, XO_3$
Answer
$H_2X$:$\text{Element/ Ion}\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \text{X}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ 2$
To stabilize the compound with hydrogen, two atoms of X are required. Therefore, the valency of X in this compound will be two.
$CX_2$:
$\text{Element/ Ion}\ \ \ \ \ \ \ \text{C}\ \ \ \ \ \ \text{X}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\ \ \ \ \ \ 2$
To stabilize the compound with hydrogen, two atoms of X are required. Thus, the valency of X in this compound is two.
$XO_2$_:
$\text{Element/ Ion}\ \ \ \ \ \ \ \text{X}\ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\ \ \ \ \ \ -2$
To stabilize this compound with oxygen, four atoms of X are required. Thus the valency of X in this compound if four.
$XO_3$:
$\text{Element/ Ion}\ \ \ \ \ \ \ \text{X}\ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\ \ \ \ \ \ -2$
To stabilize this compound with oxygen, six atoms of X are required. Thus, the valency of X in this compound is six.
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Question 85 Marks
State the various postulates of Dalton’s atomic theory of matter.
Answer
Postulates of Dalton's atomic theory:
  1. All the matter is made up of very small particles called 'atoms'.
  2. Atoms cannot be divided.
  3. Atoms can neither be created nor be destroyed.
  4. Atoms are of various kinds. There are as many kinds of atoms as are elements.
  5. All the atoms of a given element are identical in every respect, having the same mass, size and chemical properties.
  6. Atoms of different elements differ in mass, size and chemical properties.
  7. The 'number' and 'kind' of atoms in a given compound is fixed.
  8. During chemical combination, atoms of different elements combine in small whole numbers to form compounds. Atoms of the same elements can combine in more than one ratio to form more than one compound.
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Question 95 Marks
Write the formulae of the following compounds. Also name the elements present in them.
  1. Water.
  2. Ammonia.
  3. Methane.
  4. Sulphur dioxide.
  5. Ethanol.
Answer
  1. Water: $H_2O$; Elements present are Hydrogen and Oxygen.
  2. Ammonia: $NH_3$; Elements present are Nitrogen and Hydrogen.
  3. Methane: $CH_4$; Elements present are Carbon and Hydrogen.
  4. Sulphur dioxide: $SO_2$; Elements present are Sulphur and Oxygen.
  5. Ethanol: $C_2H_5OH$; Elements present are carbon, hydrogen and oxygen.
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Question 105 Marks
One of the forms of a naturally occurring solid compound $P$ is usually used for making the floors of houses. On adding a few drops of dilute hydrochloric acid to $P$, brisk effervescence are produced. When $50\ g$ of reactant $P$ was heated strongly, than $22\ g$ of a gas $Q $and $28\ g$ of a solid $R$ were produced as products. Gas $Q$ is the same which produced brisk effervescence on adding dilute HCl to $P$. Gas $Q$ is said to cause global warming whereas solid $R$ is used for white-washing.
a. What is (i) solid $P$ (ii) gas $Q$, and (iii) solid $R$.
b. What is the total mass of $Q$ and $R$ obtained from $50\ g$ of $P$ ?
c. How does the total mass of $Q$ and $R$ formed compare with the mass of $P$ taken?
d. What conclusion do you get from the comparison of masses of products and reactant?
e. Which law of chemical combination is illustrated by the example given in this problem?
Answer
$\text{P(solid)}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ \ }\text{Q(gas)}+\text{R(solid)}\\ \ \ ^{50\text{gm}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{22\text{gm}}\ \ \ \ \ \ \ \ \ \ ^{28\text{gm}}$
Solid $P$: Calcium Carbonate $(CaCO_3)$
Gas $Q$: Carbon dioxide $(CO_2)$
Solid $R$: Calcium oxide $(CaO)$​​​​​​​​​​​​​​
  1. Total mass of $Q$ and $R = 22gm + 28gm = 50gm$
  2. Total mass of $Q$ and $R (50\ gm)$ is equal to mass of reactant $(50\ gm)$.
  3. The law of conservation of mass is followed, i.e. total mass of product is equal to mass of reactant.
  4. Law of conservation of mass is illustrated by the example.
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Question 115 Marks
A liquid compound X of molecular mass 18u can be obtained from a number of natural sources. All the animals and plants need liquid X for their survival. When an electric current is passed through 200 grams of pure liquid X under suitable conditions, then 178 grams of gas Y and 22 grams of gas Z are produced. Gas Y is produced at the positive electrode whereas gas Z is obtained at the negative electrode. Moreover, gas Y supports combustion whereas gas Z burns itself causing explosions.
  1. Name (i) liquid X (ii) gas Y, and (iii) gas Z
  2. What is the ratio to the mass of element Z to the mass of element Y in the liquid X?
  3. Which law of chemical combination is illustrated by this example?
  4. Name two sources of liquid X.
  5. Gate an important use of Y in our life.
Answer
$\text{X(liquid)}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }\text{Y(gas)}+\text{Z(gas)}\\ \ ^{200\text{gm}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{178\text{gm}}\ \ \ \ \ \ \ \ ^{22\text{gm}}$
  1. Liquid X: Water.
Gas Y: Oxygen.

Gas Z: Hydrogen.
  1. $\frac{\text{mass of Z}}{\text{mass of Y}}=\frac{22\text{gm}}{178\text{gm}}=1\ :\ 8$
  2. Law of constant proportion is illustrated by this example.
  3. Two sources of liquid X– Sea, Well.
  4. Gas Y(oxygen) is necessary for breathing.
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Question 125 Marks
How many grams of magnesium will have the same number of atoms as $6$ grams of carbon?
$(Mg = 24\ u; C = 12\ u)$
Answer
Mass of carbon $=6 g$
We know that equal moles of all substances contain equal number of atoms ( $1$ mole of all substances contains $6.022$ $\times 10^{23}$ atoms).
The first step will be to convert $6\ g$ carbon into moles.
$1$ mole of carbon $=12 g$
$12 g$ carbon = $1 $ mole
Therefore, $6\ g$ carbon $=0.5$ moles
Since, equal moles of all substances contain equal number of atoms, $0.5$ moles of carbon will have the same number of atoms as present in $0.5$ moles of magnesium.
The next step will be to find out the mass of $0.5$ moles of magnesium in grams.
$1$ mole of magnesium $=24 g$
$0.5$ moles of magnesium $=24 \times 0.5=12 g$
Hence, $12\ g$ magnesium will contain the same number of atoms as present in $6\ g$ carbon.
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Question 135 Marks
What weight of oxygen gas will contain the same number of molecules as $56\ g$ of nitrogen gas?
$(O = 16u; N = 14u)$
Answer
Mass of nitrogen gas $=56 g$
We know that equal moles of all substances contain equal number of molecules. ( $1$ mole of all substances contains $6.022 \times 10^{23}$ molecules).
The first step will be to convert $56\ g$ nitrogen gas into moles.
$1$ mole nitrogen gas $N _2=28 g$
$28\ g$ nitrogen $=1 mole$
$56\ g$ nitrogen $=2$ moles
Since, equal moles of all substances contain equal number of molecules, $2$ moles of nitrogen will have the same number of molecules as 2$ moles of oxygen.$
The next step will be to find the mass of $2$ moles of oxygen in grams.
$1$ mole of oxygen $=32 g$
$2$ moles of oxygen $=32 \times 2=64 g$
Hence, $64\ g$ oxygen will contain the same number of molecules as $56\ g$ nitrogen gas.
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Question 145 Marks
State the law of conservation of mass. Give one example to illustrate this law.
Answer
Law of conservation of mass by LAVOISIER states that: "Mass can neither be created nor be destroyed in a chemical reaction".
So, in a chemical reaction, the total mass of reactants must be equal to the total mass of products.
For example: When calcium carbonate is heated, a chemical reaction takes place to form calcium oxide and calcium carbonate.
If $100\ gms$ of calcium carbonate is decomposed completely, then $56\ gms$ of calcium oxide and $44\ gms$ of carbon dioxide are formed.
In the above example: The total mass of products $= 56\ gms (CaO) + 44gms (CO_2) = 100gms.$
As total mass of products is equal to the total mass of reactant so, the law of conservation of mass is satisfied.
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Question 155 Marks
Which contains more molecules, $10\ g$ of sulphur dioxide $(SO_2)$ or $10\ g$ of oxygen $(O_2)$?
(Atomic masses: $S = 32\ u; O = 16\ u$)
Answer
Given,
Mass of sulphur dioxide $=10 g$
Molar mass of sulphur dioxide $=(1 \times 32)+(2 \times 16)=64 g$
We know that $64\ g$ of sulphur dioxide contain $6.022 \times 10^{23}$ molecules.
Number of molecules in $64\ g$ sulphur dioxide $=6.022 \times 10^{23}$
Number of molecules in $10\ g$ sulphur dioxide $=\frac{10 \times 6.022 \times 10^{23}}{64}$
Hence, the number of molecules in $10\ g$ sulphur dioxide is $9.4 \times 10^{22}$
Given,
Mass of oxygen $=10 g$
Molar mass of oxygen molecule $=(2 \times 16)=32 g$
Number of molecules in $32\ g$ oxygen $=6.022 \times 10^{23}$
Number of molecules in $10\ g$ oxygen $=\frac{10 \times 6.022 \times 10^{23}}{32}$
Hence, the number of molecules in $10\ g$ oxygen is $18.8 \times 10^{22}$
It is clear that the number of atoms in $10\ g$ oxygen is greater than that in $10\ g$ sulphur dioxide.
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Question 165 Marks
The mass of one molecule of a substance is $4.65 \times 10^{-23} g$. What is its molecular mass? What could this substance be?
Answer
Given,
Mass of one molecule of a substance $=4.65 \times 10^{-23}$
We know that the mass of a molecule of a substance is equal to the mass of 1 mole of that substance. 1 mole of a substance consists of $6.022 \times 10^{23}$ molecules; therefore, the mass of a molecule of a substance is equal to the mass of $6.022 \times 10^{23}$ molecules.
Therefore, to calculate the molecular mass of the substance, we need to calculate the mass of $6.022 \times 10^{23}$ molecules of that substance.
Mass of 1 molecule of the substance $=4.65 \times 10^{-23}$
Mass of $6.022 \times 10^{23}$ molecules of the substance $=6.022 \times 10^{23} \times 4.65 \times 10^{-23}=28 g$
Hence, molecular mass of the substance is $28\ u$ .
Nitrogen has a molecular mass of $28\ u$ , therefore, the substance is nitrogen.
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Question 175 Marks
The mass of one atom of an element $X$ is $2.0 \times 10^{-23}g.$
  1. Calculate the atomic mass of element $X.$
  2. What could element $X$ be?
Answer
Given,
Mass of an atom of element $X=2.0 \times 10^{-23} g$
We know that the mass of an atom of a substance is equal to the mass of $1$ mole of that substance.
$1$ mole of a substance consists of $6.022 \times 10^{23}$ atoms;
therefore, the mass of an atom of a substance is equal to the mass of $6.022 \times 10^{23}$ atoms.
Therefore, to calculate the atomic mass of element $X$,
we need to calculate the mass of $6.022 \times 10^{23}$ atoms of $X$. Mass of $1$ atom of $X=2.0 \times 10^{-23}$
Mass of $6.022 \times 10^{23}$ atoms of $X=6.022 \times 10^{23} \times 2.0 \times 10^{-23}=12 u$
Hence, the atomic mass of element $X$ is $12 u$.
Carbon has an atomic mass of $12\ u$ ; therefore, element $X$ is carbon.
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Question 185 Marks
Name the following compounds. Also write the symbols/ formulae of the ions present in them:
$CuSO_4$
$(NH_4)_2SO_4$
$Na_2O$
$Na_2CO_3$
$CaCl_2$
Answer
a. $CuSO _4$ : Copper sulphate; $Cu ^{+}{ }_2$ and $SO _4{ }^{2-}$
b. $\left( NH _4\right) 2 SO _4$ : Ammonium sulphate; $NH _4^{+}$and $SO _4{ }^{2-}$
c. $Na _2 O$ : Sodium oxide; $Na ^{+}$and $O _{2-}$
d. $Na _2 CO _3$ : Sodium carbonate; $Na ^{+}$and $CO _4{ }^{2-}$
e. $CaCl _2$ : Calcium chloride; $Ca ^{2+}$ and $Cl ^{-}$
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Question 195 Marks
If $1$ gram of sulphur dioxide contains $x$ molecules, how many molecules will be present in $1$ gram of oxygen?
$(S = 32\ u; O = 16\ u)$
Answer
Given, Mass of sulphur dioxide $=1 g$ We know that equal moles of all substances contain equal number of molecules. ( $1$ mole of all substances contains $6.022 \times 10^{23}$ molecules). Therefore, the ratio of molecules in sulphur dioxide and oxygen will be same as the ratio of their moles.
Number of moles in $1 g$ of sulphur dioxide:
$1$ mole of $SO _2=(1 \times 32)+(2 \times 16)=64 g 64 g$ of $SO _2=1 mole$ of $SO _2$ Therefore, $1\ g$ of $SO _2=\frac{1}{64}$ moles of $SO _2$ Now, $\frac{1}{64}$ moles of $SO _2$ contain $x$ molecules (given). Since equal number of molecules are present in equal moles of all substances,
$\frac{1}{64}$ moles of oxygen will also contain $x$ number of molecules.
Number of moles in $1 g$ of oxygen:
$1$ mole of $O _2=2 \times 16=32 g 32 g$ of $O _2=1$ mole of $O _2$ Therefore, $1\ g$ of $O _2=\frac{1}{32}$ moles of $O _2$ We have already stated that $\frac{1}{64}$ moles of oxygen contain $x$ molecules. Therefore, $\frac{1}{32}$ mole of oxygen will contain $=\frac{ x \times 64}{32}=2 x$
Hence, $1\ gm$ of oxygen consists of $2\ x$ number of molecules.
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Question 205 Marks
The atom of an element $X$ contains $17$ protons, $17$ electrons and $18$ neutrons whereas the atom of an element $Y$ contains $11$ protons, $11$ electrons and $12$ neutrons.
  1. What type of ion will be formed by an atom of element $X$? Write the symbol of ion formed.
  2. What will be the number of (i) protons (ii) electrons, and (iii) neutrons, in the ion formed from $X$?
  3. What type of ion will be formed by an atom of element $Y$? Write the symbol of ion formed.
  4. What will be the number of (i) protons (ii) electrons, and (iii) neutrons, in the ion formed from $Y$?
  5. What is the atomic mas of (i) $X$, and (ii) $Y$?
  6. What could the elements $X$ and $Y$ be?
Answer
  1. Element $X$ consists of $17$ protons and electrons and $18$ neutrons. Element $X$ will readily accept an electron to form a negatively charged anion.Symbol of the negatively charged $X$ is $X^-$.
  2. In the ion formed by $X$:
  • Number of protons = $17$
  • Number of electrons = $18$
  • Number of neutrons = $18$
  1. Element $Y$ consists of $11$ protons and electrons and $12$ neutrons. Element $Y$ will readily lose an electron to form a positively charged cation. Symbol of the positively charged $Y$ is $Y^+$.
  2. In the ion formed by $Y$:
  • Number of protons = $11$
  • Number of electrons = $10$
  • Number of neutrons = $12$
  1.  
  • Atomic mass of $X$ = number of protons of $X$ + number of neutrons of $X$ $= 17 + 18 = 35\ u$
  • Atomic mass of $Y$= number of protons of $Y$ + number of neutrons of $Y$ $= 11 + 12 = 23\ u$
  1. Atomic number of $X$ is $17$; therefore, element $X$ is chlorine.
Atomic number of $Y$ is $11$; therefore, element $Y$ is sodium.
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Question 215 Marks
Give the formulate of the compounds formed from the following sets of elements:
  1. Calcium and fluorine.
  2. Hydrogen and sulphur.
  3. Nitrogen and hydrogen.
  4. Carbon and chlorine.
  5. Sodium and oxygen.
  6. Carbon and oxygen.
Answer
  1. $\ \text{Element}:\ \ \ \ \ \text{Ca}\ \ \ \ \ \ \ \ \ \ \text{F}\\ {\text{Valencies}}:\ \ +2\ \ \ \ \ \ -2$
Thus, the resulting compound is $CaF_2$
  1. $\ \text{Element}:\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \text{S}\\ {\text{Valencies}}:\ \ +1\ \ \ \ \ -2$
Thus, the resulting compund is $H_2S$
  1. $\ \text{Element}:\ \ \ \ \ \ \ \text{N}\ \ \ \ \ \ \ \ \ \ \text{H}\\ {\text{Valencies}}:\ \ -3\ \ \ \ \ \ +1$
Thus, the resulting compound is $NH_3$​​​​​​​
  1. $\ \text{Element}:\ \ \ \ \ \ \text{C}\ \ \ \ \ \ \ \ \ \ \ \text{Cl}\\ {\text{Valencies}}:\ \ +4\ \ \ \ \ \ -1$
Thus, the resulting compound is $CCl_4$​​​​​​​
  1. $\ \text{Element}:\ \ \ \ \ \text{Na}\ \ \ \ \ \ \ \ \ \ \text{O}\\ {\text{Valencies}}:\ \ +1\ \ \ \ \ \ -2$
Thus, the resulting compound is $Na_2O$​​​​​​​
  1. $\ \text{Element}:\ \ \ \ \ \ \ \text{C}\ \ \ \ \ \ \ \ \ \ \text{O}\\ {\text{Valencies}}:\ \ +4\ \ \ \ \ \ -2$
Thus, the resulting compound is $CO_2$​​​​​​​
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