Explanation:
Water Content of Epidermal Cells
Temperature: Increase in the temperature causes stomata to open
Answer:
This part require data such as Avogadro's number and the molar mass of water. But first, let's find the mass of water in the specified volume by making use of the density formula:
Density = mass/volume
1 g/mL = Mass/70 mL
Mass = 70 g
Each water contains 18 grams per mole, and each mole contains 6.022×10²³ molecules of water. Thus,
70 g * 1mole/18 g * 6.022×10²³ molecules/mole = 2.342×10²⁴ molecules of water
Explanation:
Answer:
3–methyl–2–butanol
Explanation:
To name the compound, we must:
1. Identify the functional group.
2. Give the functional group of the compound the lowest possible count.
3. Locate the longest continuous carbon chain. This gives the parent name of the compound.
4. Identify the substituent group attached.
5. Give the substituent group the lowest possible count.
6. Combine the above to get the name of the compound.
Now, let us obtain the name of the compound.
1. The functional group of the compound is Alcohol i.e —OH.
2. The functional group is located at carbon 2.
3. The longest continuous carbon chain is carbon 4 i.e butane. But the presence of the functional group i.e OH will replace the –e in butane with –ol. Therefore, the compound is butanol.
4. The substituent group attached is methyl i.e CH3.
5. The substituent group is located at carbon 3.
6. Therefore, the name of the compound is:
3–methyl–2–butanol.
80.1 g* (1 cm^(3)/ 2.70 g)= 29.666666... cm^(3).
The calculated volume should have 3 significant figures, so the final answer is 2.97*10^(1) cm^(3)
Hope this helps~
<h3>
Answer:</h3>
1.25 moles (R.T.P.) or 1.34 moles (S.T.P.)
<h3>
Explanation:</h3>
- 1 mole of a gas occupies a volume of 24 liters at room temperature and pressure (R.T.P.)
- On the other hand, 1 mole of a gas will occupy 22.4 Liters at standard temperature and pressure (S.T.P.)
Therefore, at R.T.P.
30.0 Liters will be equivalent to;
= 30.0 L ÷ 24 L
= 1.25 moles
At S.T.P
30.0 Liters will be equivalent to;
= 30.0 L ÷ 22.4 L
= 1.34 moles
Thus, 30.0 L of helium gas are equivalent to 1.25 moles of He at R.T.P. and 1.34 moles at S.T.P.
