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Maurinko [17]
2 years ago
14

For the reaction: CaO + SO3 → CaSO4, how many moles of CaSO4 are produced

Chemistry
1 answer:
Allisa [31]2 years ago
7 0

Answer:

here hope it helps can you mark me brainliest please

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Calculate the mass defect for the formation of phosphorus-31. The mass of a phosphorus-31 nucleus is 30.973765 amu. The masses o
Nata [24]

<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399

<u>Explanation:</u>

Mass defect is defined as the difference in the mass of an isotope and its mass number.

The equation used to calculate mass defect follows:

\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M

where,

n_p = number of protons

m_p = mass of one proton

n_n = number of neutrons

m_n = mass of one neutron

M = mass number of element

We are given:

An isotope of phosphorus which is _{15}^{31}\textrm{P}

Number of protons = atomic number = 15

Number of neutrons = Mass number - atomic number = 31 - 15 = 16

Mass of proton = 1.00728 amu

Mass of neutron = 1.00866 amu

Mass number of phosphorus = 30.973765 amu

Putting values in above equation, we get:

\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399

Hence, the mass defect for the formation of phosphorus-31 is 0.27399

8 0
2 years ago
Help I’m so confused
ira [324]

Answer:

Me too.  What is this for? A Lab. You are missing some kind of key info bud.

Explanation:

4 0
3 years ago
Read 2 more answers
Which of the following could not happen to an igneous rock?
galben [10]
I could not break unless you hit it with a sledge hammer
4 0
3 years ago
The units used to measure heat are?
Akimi4 [234]

Answer&Explanation:

The unit used to measure Heat is joule equal to that of Energy and is abbreviated as (J)

7 0
3 years ago
The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse
Umnica [9.8K]
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
5 0
2 years ago
Read 2 more answers
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