Answer:
The reaction between magnesium and sulphuric acid is given as :
Mg(s) + H2SO4(aq) --> MgSO4(aq) + H2(g)
Explanation:
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There are different formula you need to keep in mind when solving for [OH-]
Given that pH = 6.10
pH + pOH = 14
6.10 + pOH = 14
pOH = 7.9
[OH-] = 10^(-pOH)
[OH-] = 10^(-7.9)
[OH-] = 0.000000013
[OH-] = 1.3 x 10^-8
<h2>
<u>Answer: [OH-] = 1.3 x 10^-8</u></h2>
Answer:
<u>[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2</u>
Explanation:
2H2S(g)⇋2H2(g)+S2(g)2H2S(g)⇋2H2(g)+S2(g)
The equilibrium constant expression in terms of concentrations is:
Kc=<u>[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2</u><u>.</u>
Answer:

Explanation:
We want to convert from moles to grams, so we must use the molar mass.
<h3>1. Molar Mass</h3>
The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).
We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.
- Phosphorus (P): 30.973762 g/mol
- Iodine (I): 126.9045 g/mol
Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.
- I₃: 126.9045 * 3=380.7135 g/mol
- PI₃: 30.973762 + 380.7135 = 411.687262 g/mol
<h3>2. Convert Moles to Grams</h3>
Use the molar mass as a ratio.

We want to convert 3.14 moles to grams, so we multiply by that value.

The units of moles of PI₃ cancel.


<h3>3. Round</h3>
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.
The 2 in the ones place tells us to leave the 9.

3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>
Answer:
The overview of the subject is outlined underneath in the summary tab.
Explanation:
- The molar ratio seems to be essentially a balanced chemical equilibrium coefficient that implies or serves as a conversion factor for the product-related reactants.
- This ratio just says the reactant proportion which reacts, but not the exact quantity of the reacting product. Consequently, the molar ratio should only be used to provide theoretical instead of just a definite mass ratio.