Answer: 0.5 moles
Explanation:
Cr2(SO3)2 is the chemical formula for chromium sulphate.
Given that,
Amount of moles of Cr2(SO3)2 (n) = ?
Mass of Cr2(SO3)2 in grams = 128.9g
For molar mass of Cr2(SO3)2, use the atomic masses:
Chromium, Cr = 52g;
Sulphur, S = 32g;
Oxygen, O = 16g
Cr2(SO3)2 =
(52g x 2) + [(32g + 16g x 3) x 2]
= 104g + [(32g + 48g) x 2]
= 104g + [80g x 2]
= 104g + 160g
= 264g/mol
Since, n = mass in grams / molar mass
n = 128.9g / 264g/mol
n = 0.488 mole [Round the value of n to the nearest tenth which is 0.5
Thus, there are 0.5 moles in 128.9 grams of Cr2(SO3)2
Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M 
solution.
Explanation:
Given: Molarity of solution = 2.0 M
Volume of solution = 250 mL
Convert mL int L as follows.
Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given solution is as follows.
Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M solution.
Answer:
a. Volatile.
b. Air foil.
c. Sash.
d. Work surface.
Explanation:
In science, matter can be defined as anything that has mass and occupies space. Any physical object that is found on earth is typically composed of matter. Matter are known to be made up of atoms and as a result has the property of existing in states.
Generally, matter exists in three (3) distinct or classical phases and these are;
I. Solid.
II. Liquid.
III. Gas.
Matching the terms with their respective definition, we have;
a. Volatile: A characteristic that describes substances that evaporate readily, producing large amounts of vapors.
b. Air foil: the front vent of a fume hood, which helps maintain proper air circulation.
c. Sash: the glass panel in front of the fume hood that shields the user from fumes and other hazards.
d. Work surface: the horizontal, flat area of a fume hood upon which experiments are carried out.
3. B
4. A
5.A
6. A (i think)
Hope this helps
C₄H₉OH + HBr = C₄H₉Br + H2O
Δmole of alcohol gives 1 mole of bromobutanol
HBr is in excess, so the yield of the product is limited by the alcohol
Wt. of 1 butanol = 18
Molar mass of the butanol = 74.12 g/mole
Moles of the alcohol = 1/74.12 = 0.01349 moles
So, moles of bromobutane = 0.01349 moles
Molar mass of C₄H₉Br = 137.018 g/moles
So, theoretical mass of bromobutane is = 0.01349 × 137.0.18
= 1.85 g
