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# Tutor profile: Dirk M.

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Dirk M.
University teaching assistant for 2 years, Currently teaching American Middle/High School Science related subjects, for the past 2 years.
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## Questions

### Subject:Chemistry

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Question:

In the following reaction: \$\$N_2\$\$ + \$\$3H_2\$\$ = \$\$2NH_3\$\$ How many grams of Hydrogen is needed to completely react with 7 grams of Nitrogen?

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Dirk M.

From the Periodic table, we find that 1 mole of \$\$N_2\$\$ has a mass of 28 g. i.e. ( 2 N x 14.0 g/mole = 28 g/mole \$\$N_2\$\$) We also know that 1 mole of \$\$H_2\$\$ is 2.02 g/mole i.e. (2 H x 1.01 g/mole = 2.02 g/mole \$\$H_2\$\$) The mole ration from the balanced equation shows: 1:3:2 \$\$N_2\$\$: \$\$H_2\$\$: \$\$NH_3\$\$ Step 1: Convert the mass of Nitrogen to moles using the mole-mass conversion equation (Mole = mass/molar mass) X mole \$\$N_2\$\$ = 7 g \$\$N_2\$\$ / 28 g/mole \$\$N_2\$\$ = 0.25 mol \$\$N_2\$\$ Step 2: Using the mole ratio from the balanced equation, 1:3 \$\$N_2\$\$: \$\$H_2\$\$, we can convert the know number of \$\$N_2\$\$ moles to the required number of \$\$H_2\$\$ moles. 0.25 mole \$\$N_2\$\$ x (3 mole \$\$H_2\$\$) / (1 mole \$\$N_2\$\$) = 0.75 mole \$\$H_2\$\$ We now know the number of moles of \$\$H_2\$\$ needed to completely react with \$\$N_2\$\$ Step 3: Convert the number of moles of \$\$H_2\$\$ to mass. 0.75 mole \$\$H_2\$\$ x 2.02 g/mole \$\$H_2\$\$ = 1.51 grams of \$\$H_2\$\$ A quicker way to complete this calculation would be to do all calculations in one step: Our ratio is: 7 g \$\$N_2\$\$ / 28 g/mole \$\$N_2\$\$ = (X)/(3 moles)(2.02 g/mole \$\$H_2\$\$). Cross multiply and the answer is 1.52 g \$\$H_2\$\$. {X = (7 g \$\$N_2\$\$)(3 moles)(2.02 g/mole \$\$H_2\$\$)/ 28 g/mole \$\$N_2\$\$ = 1.51 g \$\$H_2\$\$}

### Subject:Basic Chemistry

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Question:

Propane is a gas used for cooking and heating. Use the mole concept to calculate the number of atoms in 2.12 mol of propane (\$\$C_3H_8\$\$).

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Dirk M.

1) Analyze: List the knows and unknown. Knowns: Number of moles = 2.12 mol \$\$C_3H_8\$\$ 1 mol \$\$C_3H_8\$\$ = 6.02 X \$\$10^{23}\$\$ molecules \$\$C_3H_8\$\$ ( Avogadro’s numbers) 1 mol \$\$C_3H_8\$\$ = 11 atoms (3 Carbon atoms and 8 Hydrogen atoms) Unknown: The number of atoms in 2.12 mol \$\$C_3H_8\$\$ 2) Calculate: solve the unknown. (6.02 X \$\$10^{23}\$\$ molecules \$\$C_3H_8\$\$)/(1 mol \$\$C_3H_8\$\$) (conversion factor from mole to molecule) (11 atoms) / (1 molecule \$\$C_3H_8\$\$) (conversion factor from molecules to atoms) (2.12 mol \$\$C_3H_8\$\$) x (6.02 X \$\$10^{23}\$\$ molecules \$\$C_3H_8\$\$)/(1 mol \$\$C_3H_8\$\$) x (11 atoms) / (1 molecule \$\$C_3H_8\$\$) Therefore, We are left with the following: 2.12 x 6.02 X \$\$10^{23}\$\$ x11 = 1.4 x \$\$10^{25}\$\$ Atoms in 2.12 mol \$\$C_3H_8\$\$

### Subject:Biology

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Question:

Allele Frequency The Hardy-Weinberg principle can be used to predict the frequencies of certain genotypes if you know the frequency of other genotypes. Imagine, for example, that you know of a genetic condition, controlled by two alleles S and s, that follows the rule of simple dominance at a single locus. The condition affects only homozygous recessive individuals. (The heterozygous phenotype shows no symptoms.) The population you are studying has a population size of 10,000, and there are 36 individuals affected by the condition (\$\$q^2\$\$ = 0.0036). Based on this information, use the Hardy-Weinberg equations to answer the following questions: 1. Calculate: What are the frequencies of the S and s alleles? 2. Calculate: What are the frequencies of the SS, Ss, and ss genotypes? 3. Calculate: What percentage of people, in total, are likely to be carrying the s allele, whether or not they know it?

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Dirk M.

The Hardy-Weinberg principle states that allele frequencies in a population should remain constant unless one or more factors cause those frequencies to change. The principle makes predictions like Punnett squares- but for populations, not individuals. Suppose that there are two alleles for a gene: A (dominant) and a (recessive). A cross of these alleles can produce three possible genotypes: AA, Aa, and aa. The frequencies of genotypes in the population can be predicted by these equations, where p and q are the frequencies of the dominant and recessive alleles: In symbols: \$\$p^2\$\$ + 2pq +\$\$q^2\$\$ = 1 and p + q = 1 In words: (frequency of AA) + (frequency of Aa) + (frequency of aa) = 100% And (frequency of A) + (frequency of a) = 100% 1. Calculate: What are the frequencies of the S and s alleles? From the question we know that \$\$q^2\$\$ = 0.0036 therefore, we can calculate q: q = \$\$sqrt(0.0036)\$\$ = 0.06 x 100 = 6% - According to the Hardy-Weinberg principle, we have the equation: p + q = 1. Therefore: p + 0.06 = 1 and p = 1 – 0.06 = 0.94 or = 94% Since q stands for homozygous recessive individuals, and p stands for homozygous dominant individuals, the frequency of the s allele is 0.06 or (6%) and the frequency of the S allele is 0.94 or (94%) 2. Calculate: What are the frequencies of the SS, Ss, and ss genotypes? - According to the Hardy-Weinberg principle, we have another equation to identify the frequency of other genotypes: \$\$q^2\$\$ + 2qp + \$\$p^2\$\$ = 1 \$\$(0.94)^2\$\$ + 2 (0.94) (0.06) + \$\$(0.06)^2\$\$ = 1 Therefore: \$\$p^2\$\$ = 0.8836 = 88.36% \$\$q^2\$\$ = 0.0036 = 0.36% 2pq = 0.1128 = 11.28% Since \$\$p^2\$\$ stands for homozygous dominant individuals, it represents the SS genotypes. Since \$\$q^2\$\$ stands for homozygous recessive individuals, it represents the ss genotypes. Since 2pq stands for heterozygous individuals, it represents the Ss genotypes. Therefore, the frequency of the SS genotypes is 88.36%, the frequency of the ss genotypes is 0.36%, and the frequency of the Ss genotypes is 11.28%. 3. Calculate: What percentage of people, in total, are likely to be carrying the s allele, whether or not they know it? There are two groups that carrying the s allele, the homozygous recessive group \$\$(q^2)\$\$ and the heterozygous group (2pq). Even though the heterozygous group does not express the s allele (recessive allele), which means it does not know it (since the dominant alleles are much stronger). Hence, the percentage of people, in total, is likely to be carrying the s allele is: \$\$q^2\$\$ + 2pq = 11.28% + 0.36% = 11.64%

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