<u>Answer:</u>
<u>For a:</u> The balanced equation is 
<u>For c:</u> The balanced equation is 
<u>Explanation:</u>
A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.
The given unbalanced equation follows:

To balance the equation, we must balance the atoms by adding 2 infront of both
and
and 3 in front of 
For the balanced chemical equation:

The given balanced equation follows:

The given equation is already balanced.
The given unbalanced equation follows:

To balance the equation, we must balance the atoms by adding 2 infront of 
For the balanced chemical equation:
The given balanced equation follows:

The given equation is already balanced.
Answer:
Element 1
Explanation:
The ionization energy is defined as the energy required to remove electrons from the atoms.
We know that the nucleus of the atom attracts the electrons, thus, bound these electrons to the atom.
This means that as the radius decreases, the force of attraction between the nucleus and the electron will increase, therefore, the energy required to remove the electron would increase (and vice-versa).
Based on the above, the atom with the smallest radius would be the atom with the largest first ionization energy.
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Answer:
Trial Number of moles
1 0.001249mol
2 0.001232mol
3 0.001187 mol
Explanation:
To calculate the <em>number of moles of tritant</em> you need its<em> molarity</em>.
Since the<em> molarity</em> is not reported, I will use 0.1000M (four significant figures), which is used in other similar problems.
<em>Molarity</em> is the concentration of the solution in number of moles of solute per liter of solution.
In this case the solute is <em>NaOH</em>.
The formula is:

Solve for the <em>number of moles:</em>

Then, using the molarity of 0.1000M and the volumes for each trial you can calculate the number of moles of tritant.
Trial mL liters Number of moles
1 12.49 0.01249 0.01249liters × 0.1000M = 0.001249mol
2 12.32 0.01232 0.01232liters × 0.1000M = 0.001232mol
3 11.87 0.01187 0.01187liters × 0.1000M = 0.001187 mol
Answer:
When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)
Explanation:
Step 1: Data given
The initial volume of the sample = 2.00 L
The initial temperature = 35 °C = 308 K
The increased volume = 5.25 L
Pressure = constant
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒ with V1 = the initial volume = 2.00 L
⇒ with T1 = the initial volume = 308 K
⇒ with V2 = the new volume = 5.25 L
⇒ with T2 = the new temperature
2.00 / 308 = 5.25 / T2
0.00649 = 5.25/T2
T2 = 5.25/ 0.00649
T2 = 808.9 K
When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)
Answer:
The percentage of carbon in a 55 % pure sample of carbon carbonate is. (Ca=40, C=12, O=16)
Explanation:
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