Explanation:
We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.
Next we will calculate how many moles of
are present in 85.00 mL of 1.500 M sulfuric acid.
As, Molarity = 
1.500 M = 
n = 0.1275 mol
Now set up and solve a stoichiometric conversion from moles of
to grams of
. As, the molar mass of
is 84.01 g/mol.
= 21.42 g
So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.
I think it might be A. energy...hope this helps!
C. Metal
You can think of aluminum cans foils
They are both malleable (to morph easily)
They both are ductile (to flatten out)
They both have luster (shiny)
And they both conduct electricity.
Answer:
B.) 117 g
Explanation:
(Step 1)
To find the mass, you need to first find the moles of NaCl using the molarity ratio.
Molarity = moles / volume (L)
2.00 M = moles / 1.0 L
2.00 = moles
(Step 2)
Now that you know the moles, you can convert it to grams using the molar mass.
Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol
Molar Mass (NaCl): 58.443 g/mol
2.00 moles NaCl 58.443 g
--------------------------- x ----------------- = 117 g NaCl
1 mole