Answer:
0.56 g
Explanation:
<em>A chemist determines by measurements that 0.020 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates.</em>
Step 1: Given data
Moles of nitrogen gas (n): 0.020 mol
Step 2: Calculate the molar mass (M) of nitrogen gas
Molecular nitrogen is a gas formed by diatomic molecules, whose chemical formula is N₂. Its molar mass is:
M(N₂) = 2 × M(N) = 2 × 14.01 g/mol = 28.02 g/mol
Step 3: Calculate the mass (m) corresponding to 0 0.020 moles of nitrogen gas
We will use the following expression.
m = n × M
m = 0.020 mol × 28.02 g/mol
m = 0.56 g
Shiny
Somewhat reactive
Standard temp. and pressure
Answer:
C. If it is tested and the evidence does not support it.
Explanation:
A hypothesis is more less a scientific guess. Before such a guess or prediction is made, empirical observations and deductions are first made. It is from the result of the observations that a hypothesis statement is made.
For a hypothesis to become widely adopted and accepted, it must be testable within the limits of the experiment as described by the proposer. When subjected to test and it agrees, the status of a hypothesis can be upgraded.
If the hypothesis is tested and evidence contrasts the result being sort for, a hypothesis will be discarded.
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
<em>Answer:</em>
- 0.052301 km have 5 significant figure
- 400 cm have 1 significant figure
- 50.0 m have 3 significant figure
- 4500.01 ml have 6 significant figure
<em>Explanation:</em>
According to rules of significant figure
0.052301 km have 5 significant figure:
- Zero to the left of the first non zero digit not significant.
- Zero between the non zero digits are significant.
<em>400 cm have 1 significant figure:</em>
- Trailing zeros are not significant in numbers without decimal points.
<em>50.0 m have 3 significant figure:</em>
- Trailing zeros are significant in numbers when there is decimal points.
<em>4500.01 ml have 6 significant figure:</em>
- Zero between the non zero digits are significant.
