Some Basic Concepts of Chemistry
Get insights from 131 questions on Some Basic Concepts of Chemistry, answered by students, alumni, and experts. You may also ask and answer any question you like about Some Basic Concepts of Chemistry
- 131
Questions
- 0
Discussions
- 0
Active Users
- 0
Followers
Have a question about Some Basic Concepts of Chemistry? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
<p><strong>Among 10? g (each) of the following elements, which one will have the highest number of atoms?</strong><br><strong>Element: Pb, Po, Pr and Pt</strong><br><strong><!-- [if !supportLineBreakNewLine]--></strong><br><!--[endif]--></p>
Number of sigma bonds are 10.
[Structure showing 10 sigma bonds in the molecule]
Upvote 
The number of significant figure in 0.00340 is __________.
The number of significant figure in 0.00340 is __________.
Number of significant figure in 0.00340 are 3.
Upvote 4.5g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is × 10-1. The value of x is_________ (Rounded off to the nearest integer)
4.5g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is × 10-1. The value of x is_________ (Rounded off to the nearest integer)
W = 4.5 g
M.W = 90
V = 250 ml
Using M =
M = 2 × 10-1 M
So, x = 2
Upvote The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0mg Na+ per mL is…………. g. (Rounded off to the nearest integer)
[Given ; Atomic weight in g mol-1 -Na : 23; N : 14; O : 16]
The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0mg Na+ per mL is…………. g. (Rounded off to the nearest integer)
[Given ; Atomic weight in g mol-1 -Na : 23; N : 14; O : 16]
Mass of Na+ in 50 ml = 70 × 50 = 3500 mg
23000 mg of Na+ is present in 85000 mg of NaNO3 (1 mole NaNO3 contains 1 mole Na+)
mg Na+ will be present in
= 12934.78 mg
= 12.93478 gm
Upvote Consider titration of NoOH solution versus 1.25M oxalic acid solution. At the end point following burette readings were obtained.
(i) 4.5mL (ii) 4.5mL
(iii) 4.44mL (iv) 4.4mL
(v) 4.44mL
If the volume of oxalic acid taken was 10.0mL then the molarity of the NaOH solution is……….M.
(Rounded-off to the nearest integer).
Consider titration of NoOH solution versus 1.25M oxalic acid solution. At the end point following burette readings were obtained.
(i) 4.5mL (ii) 4.5mL
(iii) 4.44mL (iv) 4.4mL
(v) 4.44mL
If the volume of oxalic acid taken was 10.0mL then the molarity of the NaOH solution is……….M.
(Rounded-off to the nearest integer).
Meq of NaOH = Meq H2C2O4
4.44 × N = 1.25 × 2 × 10
Upvote In the given reaction,
X + Y + 3Z XYZ3
If one mole of each of X and Y with 0.05 mol of Z gives compound XYZ3. (Given: Atomic masses of X, Y and Z are 10, 20 and 30 amu, respectively.) The yield of XYZ3 is_____________g.
(Nearest integer)
In the given reaction,
X + Y + 3Z XYZ3
If one mole of each of X and Y with 0.05 mol of Z gives compound XYZ3. (Given: Atomic masses of X, Y and Z are 10, 20 and 30 amu, respectively.) The yield of XYZ3 is_____________g.
(Nearest integer)
x + y + 3z = xyz3
1 mole 1 mole 0.05 mole
here z is limiting reagent.
mole z gives 1 mole xyz3
mass of xyz3 = n * molecular mass
=
= 0.5 * 4 =
Upvote