The traditional calcium atom has twenty protons and twenty electrons making it neutral.
The calcium in the pic is a calcium ion so the number of protons and electrons are not equivalent.
Since it's 2+ that means the ion is positively charged and for that to happen electrons are away.
So 20-2=18
There are 18 electrons
Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
Answer:
We need 226 grams of FeS
Explanation:
Step 1: Data given
Mass of FeCl2 = 326 grams
Molar mass FeCl2 = 126.75 g/mol
Step 2: The balanced equation
FeS + 2 HCl → H2S + FeCl2
Step 3: Calculate moles FeCl2
Moles FeCl2 = 326 grams / 126.75 grams
Moles FeCl2 = 2.57 moles
Step 4: Calculate moles FeS needed
For 1 mol H2S and 1 mol FeCl2 produced, we need 1 mol FeS and 2 moles HCl
For 2.57 moles FeCl2 we need 2.57 moles FeS
Step 5: Calculate mass FeS
Mass FeS = 2.57 moles * 87.92 g/mol
Mass FeS = 226 grams FeS
We need 226 grams of FeS
Answer:
Δ S = 26.2 J/K
Explanation:
The change in entropy can be calculated from the formula -
Δ S = m Cp ln ( T₂ / T₁ )
Where ,
Δ S = change in entropy
m = mass = 2.00 kg
Cp =specific heat of lead is 130 J / (kg ∙ K) .
T₂ = final temperature 10.0°C + 273 = 283 K
T₁ = initial temperature , 40.0°C + 273 = 313 K
Applying the above formula ,
The change in entropy is calculated as ,
ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )
ΔS = 26.2 J/K
Answer:
-100 kJ
Explanation:
We can solve this problem by applying the first law of thermodynamics, which states that:
where:
is the change in internal energy of a system
Q is the heat absorbed/released by the system (it is positive if absorbed by the system, negative if released by the system)
W is the work done by the system (it is positive if done by the system, negative if done on the system)
For the system in this problem we have:
W = +147 kJ is the work done by the system
Q = +47 kJ is the heat absorbed by the system
So , its change in internal energy is:
