Answer:
2.24 Liters are in 4.4 grams of CO2 at STP
E=hc/l
E=
<span><span>E=<span>(6.626 x 10-34 J s)(3.0 x 108m/s )</span><span>=2.88 x 10-19J</span></span><span>6.90 x 10-7m</span></span>
Answer:
=> 2.8554 g/mL
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 16.59 g
Volume (v) = 5.81 mL
From our question, we are to determine the density (rho) of the rock.
The formula:

Substitute the values into the formula:

= 2.8554 g/mL
Therefore, the density (rho) of the rock is 2.8554 g/mL.
Answer:
Given expression: 6(2b-4). To find the value of 6(2b-4) at b= 5, we need to substitute the b=5 in the expression, we get….Therefore, the value of 6(2b-4) is 36, when b=5.
Explanation:
Given expression: 6(2b-4). To find the value of 6(2b-4) at b= 5, we need to substitute the b=5 in the expression, we get….Therefore, the value of 6(2b-4) is 36, when b=5.