Explanation:
Cr=35.880/51=0.73≈0.7
P=21.076/31=0.67≈0.7
O=43.543/16=2.71
Divide each by 0.7 u get
Cr=1 ,P =1, O=4
Empirical formula is CrPO4
Answer:
7.96g, 33.79%
Explanation:
I'll try my best to explain the entire process behind this question ;)
From the question, you can write the reaction

Now, there are a few reasons it is like this. Oxygen is a diatomic element, meaning it doesn't and can't exist as just O. It exists as O₂. To balance, this, double the amount of water and hydrogen so there is an equal amount of each element on both sides of the reaction (4 H's, 2 O's on the reactant side, and 4 H's, 2 O's on the product side).
From this we can get a mole-to-mole ratio.
Onto the stoichiometry. Our goal in this is to convert from grams of water to grams of hydrogen, and we do so with a mole to mole ratio.

Basically, what I did was divide by water's molar mass to get moles of water, multiplied by the mole-to-mole ratio (2:2) to get moles of H2, and then multiplied by H2's molar mass to get what should be the amount of H2 produced by the reaction.
For percent yield, you can calculate it is such:

Plug the numbers in:

So, the percent yield is 33.79%
germanium has 32 protons in its nucleus
Obliviously not all rocks have holes in them
It is called vesicular texture These refers to viscles holes pores or cavities within the igneous rock.the vesicles are the result of gas expansion(bubbles).which often occur during volcanic eruptions .
Answer:
The colliding molecules need to possess certain energy which is greater than the activation energy Ea and proper orientation.
<h3>Not all collisions result in a chemical reaction because not all collisions have the have sufficient amount of energy or have the appropriate activation energy. Nor do they all have this correct orientation. Molecules need to collide in such a way that they're oriented, so the correct bonds could break in. The correct bonds conform. And then, of course, they need to have sufficient activation energy in order to initiate the breaking of the bonds.</h3>
<h3>Hope this helps please let me know If I'm wrong.</h3>