I think it’s c but I could be wrong
Answer:
![\frac{3}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{3 \sqrt{3} }{3} = \sqrt{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B%20%5Csqrt%7B3%7D%20%7D%20%20%5Ctimes%20%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B%20%5Csqrt%7B3%7D%20%7D%20%20%3D%20%20%5Cfrac%7B3%20%5Csqrt%7B3%7D%20%7D%7B3%7D%20%20%3D%20%20%5Csqrt%7B3%7D%20)
when there is a radical in the denominator, we should rationalize (mutiply the denominator and numerator by the radical) to get rid of the radical in the denominator.
Its c ok im sorry if i got the last answer wrong
Answer:
it has 0g of sugar (hopes this helps)
Answer:
3Mg(s) +2P(s) -------> Mg3P2(s) + energy
Keq= [Mg3P2]/[Mg]^3 [P]^2
Explanation:
The equation for the formation of magnesium phosphide from its elements is;
3Mg(s) +2P(s) -------> Mg3P2(s) + energy
Hence we can see that three moles of magnesium atoms combines with two moles of phosphorus atoms to yield one mole of magnesium phosphide. The equation written above is the balanced chemical reaction equation for the formation of the magnesium phosphide.
The equilibrium expression for the reaction K(eq) will be given by;
Keq= [Mg3P2]/[Mg]^3 [P]^2