Answer:
Explanation:
first to get the density of some thing you have to devide the mass by the volume so 0.00018 (divided) by 10 kg and that gives you ur answer
Answer:
(a) 0.047 g (b) 0.0016 oz (c) 0.0001 lb
Explanation:
The given mass of the sodium in the slice = 47 mg
(a) Mass has to be calculated in grams
The conversion of mg to g is shown below as:
1 mg = 10⁻³ g
So,
<u>Mass of sodium = 47 × 10⁻³ g = 0.047 g</u>
(b) Mass has to be calculated in ounces
The conversion of ounces to g is shown below as:
453.6 g = 16 oz
Or,
1 g = 16 / 453.6 oz
So,
<u>Mass of sodium = (0.047 × 16) / 453.6 oz = 0.0016 oz</u>
(c) Mass has to be calculated in pounds
The conversion of pounds to g is shown below as:
1 lb = 453.6 g
Or,
1 g = 1/ 453.6 lb
So,
<u>Mass of sodium = (0.047 × 1) / 453.6 oz = 0.0001 lb</u>
The answer is (2). If you recall Rutherford's gold foil experiment, remember that a stream of positively charged alpha particles were shot at a gold foil in the center of a detector ring. The important observation was that although most of the particles passed straight through the foil without being deflected, a tiny fraction of the alpha particles were deflected off the axis of the shot, and some were even deflected almost back to the point from which they were shot. The fact that some of the alpha particles were deflected indicated a positive charge (because same charges repel), and the fact that only a small fraction of the particles were deflected indicated that the positive charge was concentrated in a small area, probably residing at the center of the atom.
Answer:
Correct option is
B
5 liters of CH
4
(g)NO
2
at STP
No. of molecules=
22.4
5
mol=
22.4
5
×N
A
molecules
A) 5ℊ of H
2
(g)
No. of moles=
2
5
mol=
2
5
×N
A
molecules
B) 5l of CH
4
(g)
No. of moles of CH
4
=
22.4
5
mol=
22.4
5
N
A
molecules
C) 5 mol of O
2
=5N
A
O
2
molecules
D) 5×10
23
molecules of CO
2
(g)
Molecules of 5l NO
2
(g) at STP=5l of CH
4
(g) molecules at STP
Therefore, option B is correct.
Answer:
11.0 L
Explanation:
The equation for this reaction is given as;
2H2 + O2 --> 2H2O
2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O
At STP;
1 mol = 22.4 L
This means;
44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O
In this reaction, the limiting reactant is H2 as O2 is in excess.
The relationship between H2 and H2O;
44.8 L = 44.8 L
11.0 L would produce x
Solving for x;
x = 11 * 44.8 / 44.8
x = 11.0 L
