Answer:
The correct answer is is option B
b. 93.3 g
Explanation:
SEE COMPLETE QUESTION BELOW
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
a. 7.30 g
b. 93.3 g
c. 146 g
d. 150 g
e. 196 g
CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION
<span>1 ml of water weighs 1 gram so 1 liter (1000 ml) weighs 1000 grams. A 3% solution (3% = 0.03) of hydrogen peroxide (w/v) would contain 1000 grams x 0.03 or 30 grams. The chemical formula of hydrogen peroxide is H2O2 and a mole weighs 34.0147 grams/mole. So 30 grams of H2O2 divided by 34.0147 grams/mole equals 0.88 moles of H2O2. The concentration of a 3% (w/v) hydrogen peroxide solution therefore contains 30 grams of H202 (or 0.88 moles of H202) per in a liter of water (or 1000 grams H20) would thus be 0.88 moles H2O2 per liter (0.88 moles H2O2/l) .</span>
0.4649331785818406 is what 27.4 grams is converted to! You're welcome!! :)
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Actually the strength
of London dispersion forces highly depend on the total number of electrons and
the area in which they are spread. We can see clearly that iodine will have the
strongest LDF's, and hence, have the highest boiling point (and melting point).
This is also the reason why iodine is a solid at room temperature, bromine is liquid
and chlorine and fluorine are gases.
Answer:
<span>Fluorine (F2)</span>
