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Sorry I don’t know the answer but sorry about this person
6.07 grams is the theoretical yield of calcium phosphate (Ca₃(PO₄)₂).
<h3>How we calculate mass from moles?</h3>
Mass of any substance can be calculated by using moles as:
n = W/M, where
W = required mass
M= molar mass
Given chemical reaction is:
3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO₄)₂
From the stoichiometry it is clear that:
3 moles of Ca(NO₃)₂ = produce 1 mole of Ca₃(PO₄)₂
Given mass of Ca(NO₃)₂ = 96.1g
Mole of Ca(NO₃)₂ = 96.1g/164g/mol = 0.5859moles
So, 0.5859 moles of Ca(NO₃)₂ = produce 0.5859×1/3 = 0.0196 moles of Ca₃(PO₄)₂
Required mass of Ca₃(PO₄)₂ will be calculated by using moles as:
W = 0.0196mole × 310g/mole = 6.07 grams
Hence, 6.07 grams is the theoretical yield of calcium phosphate.
To know more about moles, visit the below link:
brainly.com/question/15373263
Answer :
The equilibrium concentration of CO is, 0.016 M
The equilibrium concentration of Cl₂ is, 0.034 M
The equilibrium concentration of COCl₂ is, 0.139 M
Explanation :
The given chemical reaction is:

Initial conc. 0.1550 0.173 0
At eqm. (0.1550-x) (0.173-x) x
As we are given:

The expression for equilibrium constant is:
![K_c=\frac{[COCl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCOCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
Now put all the given values in this expression, we get:

x = 0.139 and x = 0.193
We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.139
The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M
The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M
The equilibrium concentration of COCl₂ = x = 0.139 M
There are 66 neutrons in a single atom of indium-115. The atomic number of indium-115 is 49, meaning there are 49 protons. Then the atomic mass is 115, so 115-49 = 66.