Its 700 J! I just took the test so yeah.... :)
Answer:
=71.439911 u
Explanation:
We know that 1 mass of hydrogen atom = 1.00784 u
If it's measured at 72u: 72/1.00784=71.439911 u
Answer:
- Add AgNO₃ solution to both unlabeled flasks: based on solubility rules, you can predict that when you add AgNO₃ to the NaCl solution, you will obtain AgCl precipitate, while no precipitate will be formed from the NaClO₃ solution.
Explanation:
<u>1. Adding AgNO₃ to NaCl solution:</u>
- AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)
<u>2. Adding AgNO₃ to NaClO₃ solution</u>
- AgNO₃ (aq) + NaClO₃ (aq) → AgClO₃ (aq) + NaNO₃ (aq)
<u />
<u>3. Relevant solubility rules for the problem.</u>
- Although most salts containing Cl⁻ are soluble, AgCl is a remarkable exception and is insoluble.
- All chlorates are soluble, so AgClO₃ is soluble.
- Salts containing nitrate ion (NO₃⁻) are generally soluble and NaNO₃ is not an exception to this rule. In fact, NaNO₃ is very well known to be soluble.
Hence, when you add AgNO₃ to the NaCl solution the AgCl formed will precipitate, and when you add the same salt (AgNO₃) to the AgClO₃ solution both formed salts AgClO₃ and NaNO₃ are soluble.
Then, the precipiate will permit to conclude which flask contains AgCl.
<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>
