Given molecule:
CH3-CH2-CH2-CH2-CH2-CH2-CH3
- The molecule contain only C and H atoms, hence it is a hydrocarbon
- The C atoms in the linear chain are linked together by carbon-carbon single bonds, hence it is a saturated compound
- It has 7 C atoms and 16 H atoms, hence the molecular formula is C7H16
-The molecular formula can be generally represented as CnH2n+2 where n = 7. This corresponds to the general molecular formula for alkanes
The given molecule is best described as an <u>alkane</u>
Answer:Calculate The PH Of The Solution After The Addition Of The Following Amounts Of HCl. PLEASE HELP! SHOW ALL STEPS!! This problem has been solved!
Explanation:
Answer:
Option D. 0.115 M
Explanation:
The following data were obtained from the question:
Mass of CuSO4 = 36.8 g
Volume of solution = 2 L
Molar mass of CuSO4 = 159.62 g/mol
Molarity of CuSO4 =..?
Next, we shall determine the number of mole in 36.8 g of CuSO4.
This can be obtained as shown below:
Mass of CuSO4 = 36.8 g
Molar mass of CuSO4 = 159.62 g/mol
Mole of CuSO4 =.?
Mole = mass /Molar mass
Mole of CuSO4 = 36.8 / 159.62
Mole of CuSO4 = 0.23 mole
Finally, we shall determine the molarity of the CuSO4 solution as illustrated below:
Mole of CuSO4 = 0.23 mole
Volume of solution = 2 L
Molarity of CuSO4 =..?
Molarity = mole /Volume
Molarity of CuSO4 = 0.23 / 2
Molarity of CuSO4 = 0.115 M
Therefore, the molarity of the CuSO4 solution is 0.115 M.
Answer:
2.9 grams.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:
no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.
no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.
<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.074 mol of Mg produce → 0.074 mol of MgO.
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>
The mass of pentane the student should weigh out is
The density of pentane is 0.626 gcm-3
To calculate the mass of pentane following expression is used,
(Density is defined as the mass divide by volume)
Density = mass / volume
mass of pentane = Density of pentane * Volume of pentane
mass of pentane = 0.626 gcm-3 * 45.0 mL
= 28.17 g
Here the unit of mass of pentane is g,
However the unit of density is gcm-3 and unit of volume is mL i.e. cm3
Hence, Mass = gcm-3 * cm3
Mass = g
The mass of pentane the student should weigh out is 28.17g
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