Answer:
chlorine has higher ionization than carbon
Explanation:
Chlorine is only one row below carbon, but it is three columns to the right in this case the IP of chlorine would be predicted to be greater than the IP of carbon.
Scientists use the physical and chemical properties to help them identify and classify matter. These physical and chemical properties are in a macro-perspective, in which these matter contains compounds, elements and atoms. Hence, matter can be classified in various ways, <span><span>
1. </span>Atomic number either atomic mass each element has</span>
<span><span>2. </span>By substance of that matter either pure substance or mixed substance</span> <span>
3. If they cannot reduce a certain substance into a much smaller quantified atomic structure then they they’ll use (2) to identify and classify it.</span>
They often create data tables, charts, or graphs to easily communicate results.
Answer : The mass of nitric acid is, 214.234 grams.
Solution : Given,
Moles of nitric acid = 3.4 moles
Molar mass of nitric acid = 63.01 g/mole
Formula used :

Now put all the given values in this formula, we get the mass of nitric acid.

Therefore, the mass of nitric acid is, 214.234 grams.
Answer:
c = 0.377 J/g.°C
c = 0.2350 J/g.°C
J = 27.3 J
Explanation:
We can calculate the heat (Q) absorbed or released by a substance using the following expression.
Q = c × m × ΔT
where,
c: specific heat
m: mass
ΔT: change in the temperature
<em>It takes 49.0J to raise the temperature of an 11.5g piece of unknown metal from 13.0°C to 24.3°C. What is the specific heat for the metal? Express your answer numerically, in J/g.°C</em>
Q = c × m × ΔT
49.0 J = c × 11.5 g × (24.3°C - 13.0°C)
c = 0.377 J/g.°C
<em>The molar heat capacity of silver is 25.35 J/mol.°C. How much energy would it take to raise the temperature of 11.5g of silver by 10.1°C? Express your answer numerically, in Joules. What is the specific heat of silver?</em>
<em />
The molar mass of silver is 107.87 g/mol. The specific heat of silver is:

Q = c × m × ΔT
Q = (0.2350 J/g.°C) × 11.5 g × 10.1°C = 27.3 J