Answer:
The coefficient of Ca(OH)2 is 1
Explanation:
Step 1: unbalanced equation
Ca(OH)2 + HNO3 → Ca(NO3)2 + H2O
Step 2: Balancing the equation
On the right side we have 2x N (in Ca(NO3)2 ) and 1x N on the left side (in HNO3). To balance the amount of N on both sides, we have to multiply HNO3 by 2.
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + H2O
On the left side we have 4x H (2xH in Ca(OH)2 and 2x H in HNO3), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side, by 2.
Now the equationis balanced.
Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O
The coefficient of Ca(OH)2 is 1
Answer: 14.1g
Explanation:
Given that,
number of moles of SiO2 = 0.235 moles
Mass in grams = Z (let unknown value be Z)
Molar mass of SiO2 = ?
To get the molar mass of SiO2, use the atomic mass
Silicon = 28g;
Oxygen = 16g
i.e Molar mass of SiO2 = 28g + (16g x 2)
= 28g + 32g
= 60g/mol
Now, apply the formula
Number of moles = Mass / molar mass
0.235 moles = Z / 60g/mol
Z = 0.235 moles x 60g/mol
Z = 14.1 g
Thus, the mass of SiO2 is 14.1 grams.
<em>Answer:</em>
4) the one that is reduced, which is the oxidizing agent
<em>Explanation:</em>
<em>An oxidizing agent is one that causes oxidation by gaining electrons from another atom/molecule. </em>
<u>Answer:</u> The mass of nitrogen gas reacted to produce given amount of energy is 5.99 grams.
<u>Explanation:</u>
The given chemical reaction follows:
We know that:
Molar mass of nitrogen gas = 28 g/mol
We are given:
Enthalpy change of the reaction = 14.2 kJ
To calculate the mass of nitrogen gas reacted, we use unitary method:
When enthalpy change of the reaction is 66.4 kJ, the mass of nitrogen gas reacted is 28 grams.
So, when enthalpy change of the reaction is 14.2 kJ, the mass of nitrogen gas reacted will be = 
Hence, the mass of nitrogen gas reacted to produce given amount of energy is 5.99 grams.
Answer:
<h2>It makes the current viable enough to pass through an exterior wire.</h2>
Explanation:
Electrochemical cells primarily comprise of two half-cells. These half-cells assist in isolating the oxidation and reduction half-reactions. These two reactions are linked by a wire which allows the current to move from one edge to the other. The oxidation at the anode and the reduction take place at the cathode and the addition of a salt bridge helps in completing the circuit and permits the current to flow and leads to the generation of electricity.
