Answer:
B
Explanation:
One of the main problems is the soil erosion reducing the fertitlity of the soil, thus reducing it's chances of the soil helping plant growth, thus reducing biodiversity. The other selections worsen the problem.
02(g) = 0 kj/mol
<span>CO2 (g) = -393.5 kj/mol </span>
<span>H20(g) = -241.8 kj/mol </span>
<span>H total = -5094 kJ
</span>5094kJ = [8(-393.5) + 9(-241.8)] - [X + 12.5(0)]
<span>-5094 kJ = [-3148 + (-2176.2)] - [x + 0] </span>
<span>-5094 kJ = -5324.2 - x </span>
<span>add -5324.2 to -5094 </span>
<span>to get +230.2 = -x </span>
<span>move the negative to the other side </span>
<span>and you get -230 kj/mol</span>
Answer:
b). silver (Ag)
Explanation:
If you look at the periodic table, you just need to look at the atomic number of the element, because the atomic number tells you how many protons there are in the nucleus of the element.
But do be careful because some periodic tables have the molar mass at the top left corner, but the one I use has the atomic number at the top left corner, so make sure you look for the atomic number and not the molar mass.
Answer:
The molarity of the new solution is 0.72 M
Explanation:
Step 1: Data given
Volume of the original solution = 360 mL =.360 L
Molarity = 0.87 M
We add 75 mL = 0.075 L
Step 2: Calculate moles
Moles = molarity * volume
Moles = 0.87 M * 0.360 L
Moles = 0.3132 moles
Step 3: Calculate new molarity
The number of moles stays constant
Molarity = moles / volume
Molarity = 0.3132 moles / (0.36+0.075)
Molarity = 0.3132 moles / 0.435 L
Molarity = 0.72 M
The molarity of the new solution is 0.72 M
Answer:
<h2>pH = 3.9</h2><h2>pOH = 10.1</h2>
Explanation:
Since
is a weak acid to find the pH of
we use the formula

where
Ka is the acid dissociation constant
c is the concentration
From the question
Ka of
= 1.75 × 10^-5
c = 1.00 × 10-³M
Substitute the values into the above formula and solve for the pH
That's

We have the answer as
<h3>pH = 3.9</h3>
To find the pOH we use the formula
pH + pOH = 14
pOH = 14 - pH
pOH = 14 - 3.9
We have the answer as
<h3>pOH = 10.1</h3>
Hope this helps you