You multiply avogadro's number to what you were given.
8.30x10^23 * 6. 0221409x10^23
=1.357*10^25
That should be the right answer but I'm not sure. It has been awhile since I have done this.
Answer:
The heat that was used to melt the 15.0 grams of ice at 0°C is 4,950 Joules
Explanation:
The mass of ice in the beaker = 15.0 grams
The initial temperature of the ice = 0°C
The final temperature of the ice = 0°C
The latent heat of fusion of ice = 330 J/g
The heat required to melt a given mass of ice = The mass of the ice to be melted × The latent heat of fusion of ice
Therefore, the heat, Q, required to melt 15.0 g of ice = 15.0 g × 330 J/g = 4,950 J
The heat that was used to melt the 15.0 grams of ice = 4,950 Joules.
Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234
So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left
Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.
Hey there!:
HCl + MnO2 → MnCl2 + H2O + Cl2
* in HCl the oxidation state of Cl is -1 .
* on the product side the oxidation state is 0 .
* therefore Cl gains electrons .
* in MnO2 the oxidation state of Mn is +4
* in MnCl2 the oxidation state of Mn is +2
Therefore Mn loses electrons
Answer A
Hope That helps!
