Grams of Chromium(III) nitrate produced : 268.95 g
<h3>Further explanation</h3>
Given
0.85 moles of Lead(IV) nitrate
Required
grams of Chromium(III) nitrate
Solution
Reaction
Balanced equation :
<em>2Cr₂ + 3Pb(NO₃)₄ ⇒ 4Cr(NO₃)₃ + 3Pb </em>
From the equation, mol ratio of Pb(NO₃)₄ : Cr(NO₃)₃ = 3 : 4, so mol Cr(NO₃)₃
mol Cr(NO₃)₃ =
Mass of Chromium(III) nitrate (MW=238.0108 g/mol) :
mass = mol x MW
mass = 1.13 x 238.0108
mass = 268.95 g
Answer:
We need 375 milliliters of 0.100 M NaHCO3 solution
Explanation:
Step 1: Data given
Initial molarity NaHCO3 = 0.100 M
Volume prepared solution = 750.0 mL
Molarity prepared solution = 0.05 M
Step 2: Calculate initial volume
C1*V1 = C2V2
⇒with C1 = the initial concentration = 0.100 M
⇒with V1 = The initial volume = TO BE DETERMINED
⇒with C2 = the new concentration = 0.0500M
⇒with v2 = the new volume = 750.0 mL = 0.750 L
0.100 M * V1 = 0.0500 M * 0.750 L
V1 = (0.0500M * 0.750L)/0.100 M
V1 = 0.375 L = 375 mL
We need 375 milliliters of 0.100 M NaHCO3 solution
Answer:
Ketone
Explanation:
According to the description of the functional group attached to the hydrocarbon - the first and last of a chain of three carbons are each single bonded to 3 hydrogen (H)atoms; the center carbon is double bonded to an oxygen atom, O - it can be deduced that the functional group is a ketone group.
A ketone is a functional group with the structure R₂-C=O, where the two R are alkyl or phenyl groups which may be the same or different. The carbon in a double covalent bond with oxygen is called the carbonyl carbon. The two R groups are bonded to the carbonyl carbon in single covalent bonds.
In the hydrocarbon in this question, the two R groups are methyl groups, CH₃ bonded to the carbonyl carbon. The name of the hydrocarbon is propanone and it is a ketone molecule.
Answer:
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Explanation:
Answer:- d) 42 kJ of heat is released and the reaction is exothermic.
Solution:- Heat of reaction is the summation of heats of products - reactants.
![\Delta H_r_x_n=\sum [products-reactants]](https://tex.z-dn.net/?f=%5CDelta%20H_r_x_n%3D%5Csum%20%5Bproducts-reactants%5D)
From given information, the energy contained by products is 352 kJ and the energy contained by reactants is 394 kJ. Let's plug in the values in the formula:
= [352 kJ - 394 kJ]
= -42 kJ
Heat of reaction is -42 kJ. The negative sign indicates the heat is released means the reaction is exothermic.
So, the correct option is the last one, 42 kJ of heat is released and the reaction is exothermic.
