Answer:
The correct answer is C.
Explanation:
To solve this problem, we should first recall possible units for mass and volume. Mass can be represented with units of grams, kilograms, etc. Volume has units of cubic meters, cubic centimeters, etc.
Therefore, since we know that density is mass divided by volume, the only answer that makes sense here is C. grams per cubic centimeter (where per signifies that grams are being divided by cubic centimeters).
Hope this helps!
Answer:
Be. Beryllium
Explanation:
Answer: 7.8 moles of NaCl result from the complete reaction of 3.9 mol of
Explanation:
To calculate the moles :
As is the excess reagent, is the limiting reagent as it limits the formation of product.
According to stoichiometry :
1 mole of gives = 2 moles of
Thus 3.9 moles of will give= of
7.8 moles of NaCl result from the complete reaction of 3.9 mol of
Answer:
50 g NaOH
General Formulas and Concepts:
Chemistry - Solutions
Explanation:
Step 1: Define
0.500 L
2.50 M NaOH
x moles of solute
Step 2: Define conversions
Molar Mass of Na - 22.99 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H - 1.01 g/mol
Molar Mass of NaOH - 22.99 + 16.00 + 1.01 = 40 g/mol
Step 3: Find moles of solute
2.50 M NaOH = x moles NaOH / 0.500 L
x = 1.25 moles NaOH
Step 4: Convert
= 50 g NaOH
Step 5: Check
We are given 3 sig figs. Since our final answer has 2 sigs, there is no need to round.
Answer:
C.
Explanation:
Molecules with the stronger intermolecular forces are pulled tightly together to form solid at higher temperatures and that's why the freezing point is higher.
Also, molecules with the stronger intermolecular force have greater interaction between the molecules and thus on heating do not boil easily and have high boiling point also.
Thus, melting point and boiling point increases with increase in number of carbon atoms and also increase in intermolecular forces (like hydrogen bonding, if present).
Thus, the compound which is gas at room temperature is because it has least number of carbon atoms and absence of hydrogen bonding.
Among the choices, only CH3CH2CH3 (propane) is a gas at room temperature. The other compounds, CH3CH2OH (ethanol), CH3CH2CH2CH2CH2CH2CH3 (hexane), and HOCH2CH2OH (ethylene glycol) are all liquids.
Among the four compounds given, compound C, which is CH3CH2CH3 (also known as propane), is a gas at room temperature. Compound A (CH3CH2OH, or ethanol), compound B (CH3CH2CH2CH2CH2CH2CH3, or hexane), and compound D (HOCH2CH2OH, or ethylene glycol) are all liquids at room temperature.
The state of a compound at room temperature depends on factors like molecular mass and intermolecular forces. Propane has a smaller molecular mass and weaker intermolecular forces than the others, making it a gas at room temperature.
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I don't understand this
Answer: Galaxy D is what i got and it says ts is right
Explanation:
the red line on d is closer than the rest
Answer: you cannot find its new volume by using these known values and Boyle's law because the temperature does not remain constant.
Explanation:
Boyle's law states that the volume of a fixed amount of gas, at a constant temperature, varies inversely with the pressure.
So, it is a condition that the temperature does not change.
For the wheater ballon case, as it travels through the atmosphere, the temperature at different altitudes will be different.
So, you might use other equation of states, such as the combined law, which does deal with changes in the three variables: volume, pressure, and temperature.
The mathematical formulation of Boyle's law is:
pV = constanjt ⇒ p₁ V₁ = p₂ V₂, at constant T.
The mathematical formulation of the combined law of gases is:
pV/T = constant ⇒ p₁ V₁ / T₁ = p₂ V₂ / T₂, for a fixed amount of gas, then it might work for the weather ballon (if you know the initial and end temperatures).
You cannot find the new volume by using initial volume of the weather balloon and air pressure ai its initial and final altitudes and Boyle’s law because the given values are not the same. Boyle’s law holds for the pressure and volume of the GAS at constant temperature. Here you are given the air pressure outside the weather balloon not the inside of the balloon. They have different gases and so it would not apply.