When would a life scientist study a nonliving thing such as a rock or lake

Draw fischer projection for d-2-ketotetrose

koban [17]

Fischer projection for D-2-ketotetrose is in Word document below.
D-2-ketotetrose is monosaccharide, having both a ketone (a ketose) and four carbons (a tetrose). There are two ketotetroses (the enantiomers) L and D-erythrulose, this is D-erytrhrulose (1,3,4-trihydroxy-2-butanone).
The Fischer projection is two-dimensional representation of a three dimensional organic molecule.

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3 0

3 years ago

The amount of carbon dioxide (in pounds per 15,000 cubic miles) released by a certain SUV depends on its fuel efficiency accordi

Shtirlitz [24]

Answer:

30 miles per gallon.

Explanation:

See picture

When the amount of carbon is plotted based on efficiency, the following graph is obtained (see image)

when looking at the graph, it can be seen that the point of lowest emission is presented with the maximum possible efficiency, 30 miles per gallon and this value is 10400 pounds per 15000 cubic miles.

$W=32x^{2}-2080x+44000$

5 0

3 years ago

Calculate the mass of excess reagent remaining at the end of the reaction in which 90.0 g of SO2 are mixed with 100.0 g of O2

Radda [10]

We have to first write a balanced equation.
so2 + o2 -> so3

this is not balanced though. we have 3 oxygen on right and 4 on left
2so2 + o2 -> 2so3

now it is same on both sides. we have to figure out which is limiting reagent with the given amounts of reagents. we do this by comparing the ratio between them in terms of moles. we see that so2 has a coefficient of 2 and o2 has none which implies 1 and so3 has 2. this means that for every 2 moles of so2 reacting with 1 mole of o2, we get 2 moles of so3.

lets convert the given values to moles. to do this we know that molecular weight is measured in grams per mole. we are given grams and need to cancel out the grams to get moles. so the molecular weight:
so2 =32.1 + 2 * 16 = 64.1 g/mol
o2 = 2 * 16 = 32 g/mol
so3 = 32.1 + 3 * 16 = 80.1 g/mol

now to convert 90 g of 2so2 under ideal conditions.
90g / 64.1g/mol = 1.404 moles

convert this amount of moles of so2 to moles of o2. we have 2 moles of so2 to 1 of o2
1.404moles so2 / 2 moles so2 * 1 mole o2= 0.702 moles o2

so we see under ideal conditions that 90g of so2 would react with .702g of o2. lets see how many we actually have with 100g of o2
100g / 32g/mol =3.16 mol.

so we have a lot more o2 than needed. we are looking for how much is left in grams. we have to figure out how much was used. to do this convert our ideal moles of o2 into grams.
.702 moles o2 * 32g/mol = 22.5g o2

so what we startrd with (100g) minus what we needed (22.5g) is what we have left
100 - 22.5 = 77.5g o2

6 0

3 years ago

Which is the next logical step in balancing the given equation? CS2(1)+Cl2(g) CCl4(1) S2Cl2(1)

Mandarinka [93]

Reaction of current interest is:

CS2(l) + Cl2(g) → CCl4(l) + S2Cl2(l)

While balancing the chemical reaction, care must be taken that number of atoms of reactant side is equal to number of reactant on product side.

In present case, There is 1 'C' atom of both reactant and product side
There are 2 'S' atoms on both reactant and product side.
However, there are 2 Cl atoms of reactant side, but 6 Cl atoms on product side.
Hence multiplying Cl2 by 3, would equal the number of Cl atoms on both the sides.

Thus, the balanced reaction is
CS2(l) + 3 Cl2(g) → CCl4(l) + S2Cl2(l)

7 0

3 years ago

A nitrogen oxide, containing 53.85% N, acts as a vasodilator, lowering blood pressure in the human body. What is its empirical f

jenyasd209 [6]

The empirical formula of the nitrogen oxide that acts as a vasodilator and lowers human blood pressure is $N_4O_3$ . Empirical formula is the smallest whole number ratios of elements in a compound.

FURTHER EXPLANATION

To get the empirical formula from mass percent data, the following steps are followed:

1. Get the mass percent data for all the elements that make up the compound.

2. Assume that there is 100 grams of sample compound. Determine the equivalent masses of each element using the mass percent given.

3. Convert the mass of each element to number of moles.

4. Divide each calculated mole by the smallest mole value.

5. The quotient will be the subscript of the element in the compound. If a decimal is obtained, round off the answers to the nearest whole number. Some exceptions to this are the following decimals: x.25, x.33, x.50, and x.75. When these decimals are obtained, all the subscripts are multiplied by a number that will result in a whole number.

Applying the steps to the problem,

STEP 1: Get the mass percent.

Mass % of N = 53.85

Mass % of O = (100 - 53.85) = 46.15

STEP 2: Assume that 100 g sample is used and convert the mass % to mass in grams.

mass of N = 53.85% x 100 g = 53.85 g

mass of O= 46.15% x 100 g = 46.15 g

STEP 3: Convert mass to moles by dividing the given mass by the formula mass.

$moles \ of \ N \ = 53.85 \ g (\frac{1 \ mol}{14 \ g}) = 3.85 \\\\moles \ of \ O \ = 46.26 \ g \ (\frac{1 \ mol}{16 \ g}) \ = 2.89$

The moles may be written as the temporary subscripts of the elements in the compound as follows:

$N_{3.85}O_{2.89}$

STEP 4: Divide the moles by the smallest mole value.

$N_{3.85}O_{2.89}\\N_{ \frac{3.85}{2.89}} \ O _{ \frac{2.89}{2.89}} \\ N_{1.33}O_{1}\\$

STEP 5: Multiply the subscripts by a number that would give the smallest whole number ratio.

Since the decimal is 1.33 it cannot be rounded off to 1. It should be multiplied by 3 to get the smallest whole number.

NOTE: All subscripts must be multiplied by the same factor, 3.

$N_{1.33}O_{1}\\3(N_{1.33}O_{1})\\\boxed {N_{4}O_{3}}$

To check if the empirical formula is correct, calculate the mass % of each element based on the formula and compare with the given in the problem.

$mass \ \%\ = (\frac{mass \ of \ element}{mass \ of \ compound}) x 100\\\\mass \ \% \ of \ N = \ \frac{(4)(14)}{(4)(14) \ + \ (3)(16)}$

$mass \ \% \ of \ N \ = 53.85%\\mass \ \% \ of \ O \ = (\frac{(3)(16)}{(4)(14)+(3)(16)}) \ 100\\mass \ \% \ of \ O \ = 46.15%$

Since the values are the same from the given, the answer is correct.

LEARN MORE

Writing Chemical Formula brainly.com/question/4697698
Naming Compounds brainly.com/question/8968140
Mole Conversion brainly.com/question/12979299

Keywords: empirical formula, compounds

6 0

3 years ago

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