Answer:
80.1 grams
Explanation:
Find the molar mass of CH3OH first by using the periodic table values.
12.011 g/mol C + (1.008*3 g/mol H) + 15.999g/mol O + 1.008 g/mol H
=32.042 so that is the molar mass
Now that you have 2.50 moles of CH3OH, you can calculate the mass in g
2.50molCH3OH * (32.042g CH3OH / 1 mol CH3OH) = 80.105
32.042g / 1 mol is the same as 32.042 g/mol
Since there are 3 sig figs in the problem (2.50 has 3 sig figs), you round to 80.1 g CH3OH
Use the formula in terms of half life from the normal exponential functions
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N(t) = N(0) (1/2) ^ (t/thalf) </span>
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N(0) is the original quantity </span>
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N(t) = quantity remaining at time t </span>
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t is the time </span>
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thalf is half life </span>
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1/16 = (1/2)^(t/3.82) </span>
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16 = 2^(t/3.82) </span>
<span>
4 = t/3.82 </span>
<span>
t = 15.28 days
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<span>Answer: option (4) the same magnitude and the opposite sign.
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Justification:
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</span><span>1) Electrons are negative particles thar are around the nucleus of the atom (in regions called orbitals).
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2) Protons are positive particles that are inside the nuclus of the atom.
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<span>3) The nucleus of the atom has the same number of protons as electrons are in the orbitals of the atom.
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4) The atoms are neutral (neither positive nor negative) because there are the same number of electrons and protons and their charge are of the same magnitude but different sign: (+) + (-) = 0: positive + negative = neutral.</span>
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
Answer:
Theoretical yield of C6H10 = 3.2 g.
Explanation:
Defining Theoretical yield as the quantity of product obtained from the complete conversion of the limiting reactant in a chemical reaction. It can be expressed as grams or moles.
Equation of the reaction
C6H11OH --> C6H10 + H2O
Moles of C6H11OH:
Molar mass of C6H110H = (12*6) + (1*12) + 16
= 100 g/mol
Mass of C6H10 = 3.8 g
number of moles = mass/molar mass
=3.8/100
= 0.038 mol.
Using stoichoimetry, 1 moles of C6H110H was dehydrated to form 1 mole of C6H10 and 1 mole of water.
Therefore, 0.038 moles of C6H10 was produced.
Mass of C6H10 = molar mass * number of moles
Molar mass of C6H10 = (12*6) + (1*10)
= 82 g/mol.
Mass = 82 * 0.038
= 3.116 g of C6H10.
Theoretical yield of C6H10 = 3.2 g