<span>pre-1982 definition STP: 120 g/mol
post-1982 definition STP: 122 g/mol
The answer to this question depends upon which definition of STP you're using. The definition changed in 1982 from 273.15 K at 1 atmosphere to 273.15 K at 10000 pascals. As a result the molar volume of a gas at STP changed from 22.4 L/mol to 22.7 L/mol. So let's calculate the answer using both definitions and see if your text book is 35 years obsolete.
First, determine the number of moles of gas you have. Do this by dividing the volume you have by the molar volume. So
pre-1982: 0.04665 / 22.4 = 0.002082589 mol
post-1982: 0.04665 / 22.7 = 0.002055066 mol
Now divide the mass you have by the number of moles.
pre-1982: 0.250 g / 0.002082589 mol = 120.0428725 g/mol
post-1982: 0.250 g / 0.002055066 mol = 121.6505895 g/mol
Finally, round to 3 significant figures:
pre-1982: 120 g/mol
post-1982: 122 g/mol
These figures are insanely large for nitrogen gas. So let's see if our input data is reasonable. Looking up the density of nitrogen gas at STP, I get a value of 1.251 grams per liter. The value of 0.250 grams in the problem would then imply a volume of about one fifth of a liter, or about 200 mL. That is over 4 times the volume given of 46.65 mL. So the verbiage in the question mentioning "nitrogen gas" is inaccurate at best.
I see several possibilities.
1. The word "nitrogen" was pulled out of thin air and should be replaced with "an unknown"
2. The measurements given are incorrect and should be corrected.
In any case, if #1 above is the correct reason, then you need to pick the answer based upon which definition of STP your textbook is using.</span>
Answer:
Protein Concentration is 2.82mg/L
Explanation:
According to Beer-Lambert's Law, Absorbance is directly proportional to the concentration.
However, the concentration of a solution can be determined from a calibration curve, in which Absorbance is plotted on the y-axis and the Concentration on the x-axis.
Plotting the best line, the equation of line is used
y = mx + c
where y is absorbance = 0.150
m is slope = 0.0163
x is concentration
c is intercept = 0.104
inserting the values from the question
y = mx + c
0.150 = 0.0163x + 0.104
0.0163x = 0.150 - 0.104
0.0163x = 0.046
Divide both sides by 0.0163
0.0163x/0.0163 = 0.046/0.0163
x = 2.82
Concentration of protein = 2.82 mg/L
Answer:
B. High levels of nitrogen and phosphorus from agricultural runoff, sewage, and industrial pollution were flowing into the bay.
Explanation:
I just did this question! <3
Answer:
V₂ ≈416.7 mL
Explanation:
This question asks us to find the volume, given another volume and 2 temperatures in Kelvin. Based on this information, we must be using Charles's Law and the formula. Remember, his law states the volume of a gas is proportional to the temperature.
where V₁ and V₂ are the first and second volumes, and T₁ and T₂ are the first and second temperature.
The balloon has a volume of 600 milliliters and a temperature of 360 K, but the temperature then drops to 250 K. So,
- V₁= 600 mL
- T₁= 360 K
- T₂= 250 K
Substitute the values into the formula.
- 600 mL /360 K = V₂ / 250 K
Since we are solving for the second volume when the temperature is 250 K, we have to isolate the variable V₂. It is being divided by 250 K. The inverse o division is multiplication, so we multiply both sides by 250 K.
- 250 K * 600 mL /360 K = V₂ / 250 K * 250 K
- 250 K * 600 mL/360 K = V₂
The units of Kelvin cancel, so we are left with the units of mL.
- 250 * 600 mL/360=V₂
- 416.666666667 mL= V₂
Let's round to the nearest tenth. The 6 in the hundredth place tells us to round to 6 to a 7.
The volume of the balloon at 250 K is approximately 416.7 milliliters.
