For the formula shown, how many sulfur atoms are represented?

What's the process of releasing energy without oxygen

LenKa [72]

That would be anaerobic fermentation

I hope I've helped!

7 0

3 years ago

How many photons are produced in a laser pulse of 0.364 J at 477 nm?

Luda [366]

Answer:

1.00 × 10¹⁸

Explanation:

1. Calculate the energy of one photon

The formula for the energy of a photon is

E = hc/λ

h = 6.626 × 10⁻³⁴ J·s; c = 2.998 × 10⁸ m·s⁻¹

λ = 477 nm = 477 × 10⁻⁹ m Insert the values

E = (6.626 × 10⁻³⁴ × 2.998× 10⁸)/(477 × 10⁻⁹)

E = 4.165× 10⁻¹⁹ J

2. Calculate the number of photons

Divide the total energy by the energy of one photon.

No. of photons = 0.418 × 1/4.165 × 10⁻¹⁹

No. of photons = 1.00 × 10¹⁸

7 0

4 years ago

Blood alcohol concentration (BAC) is a measure of alcohol in your body, expressed as grams of alcohol per 100ml of blood. For a

ella [17]

Answer:

A 71 kg person will get a BAC of 0.05 when drinking 0.3442 L of beer

Explanation:

The mass of blood in a human body is approximately 8%, so if a person weighs 71 kg, the mass of blood would be:

71 kg * 8/100 = 5.68 kg of blood.

Using blood density, we can calculate the volume that 5.68 kg of blood occupies:

5.68 kg * $\frac{1m^{3}}{1025kg}$ = 0.0055415 m³

We convert m³ into mL, keeping the unit that we want to convert to in the numerator; and the unit that we want to convert in the denominator:

$0.0055415m^{3}*\frac{1000L}{1m^{3}} *\frac{1000mL}{1L}=5541.5mL$

Now we calculate the amount of alcohol that would be needed in the bloodstream to get a BAC of 0.05:

$\frac{5541.5mL}{100mL}*0.05g$ = 2.77 g of alcohol are needed in the bloodstream in order to have a BAC of 0.05

The amount of alcohol that needs to be ingested is higher than 2.77 g, due to the fact that only 17% of the alcohol goes into the bloodstream, so:

2.77 g $*\frac{100}{17} =$ 16.29 g of alcohol need to be ingested

Then we use the alcohol concentration of beer to calculate the volume of beer needed, using the alcohol density. First we convert the alcohol density to g/L, making sure the units that we want to convert cancel each other:

$789\frac{kg}{m^{3}}*\frac{1000g}{1kg} *\frac{1m^{3}}{1000L} =789g/L$

Now we use the density to calculate the litres of alcohol needed, keeping in mind that 16.29 g of alcohol are needed:

$16.29g*\frac{1L}{789g}=$ 0.02065 L of alcohol are needed.

Finally we calculate the litres of beer needed, keeping in mind the concentration of alcohol in beer:

$0.02065L_{alcohol}*\frac{100L_{beer}}{6L_{alcohol}} =$ 0.3442 L of beer are needed.

4 0

3 years ago

What is the empirical formula of a compound that is 7.74% H and 92.26% C? What is the molecular formula if the molar mass is 78.

Minchanka [31]

Answer:

For all these questions, we want to find the empirical and molecular formulae of various compounds given their percent composition and molar mass. The technique used to answer one of the questions can accordingly be applied to all of them.

Approaching the first question, we treat the percentages of each element as the mass of that element in a 100 g compound (as the percentages add up to 100%). So, our 100 g compound comprises 7.74 g H and 92.26 g C.

Next, we convert these mass quantities into moles. Divide the mass of each element by its molar mass:

7.74 g H/1.00794 g/mol = 7.679 mol H

92.26 g C/12.0107 g/mol = 7.681 mol C.

Then, we look for the molar quantity that's the smallest ("smaller," in this case, since there are only two), and we divide all the molar quantities by the smallest one. Here, it's a very close call, but the number of moles of H is slightly smaller than that of C. So, we divide each molar quantity by the number of moles of H:

7.679 mol H/7.679 mol H = 1

7.681 mol C/7.679 mol H ≈ 1 C/H (the value is actually slightly larger than 1, but we can treat it as 1 for our purposes).

The quotients we calculated represent the subscripts of our compound's empirical formula, which should provide the most simplified whole number ratio of the elements. So the empirical formula of our compound is C₁H₁, or just CH.

Here, it just so happens that we obtained whole number quotients. If we end up with a quotient that isn't a whole number (e.g., 1.5), we would multiply all the quotients by a common number that would give us the most simplified whole number ratio (so, if we had gotten 1 and 1.5, we'd multiply both by 2, and the empirical formula would have subscripts 2 and 3).

To find the molecular formula (the actual formula of our compound), we use the molar mass of the compound, 78.1134 g/mol. The molar mass of our "empirical compound," CH, is 13.0186 g/mol. Since our empirical formula represents the most simplified molar ratio of the elements, the molar masses of our "empirical compound" and the actual compound should be multiples of one another. We divide 78.1134 g/mol by 13.0176 g/mol and obtain 6. The subscripts in our molecular formula are equal to the subscripts in our empirical formula multiplied by 6.

Thus, our molecular formula is C₆H₆.

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As mentioned before, all the questions here can be answered following the procedure used to answer the first question above. In any case, I've provided the empirical and molecular formulae for the remaining questions below for your reference.

2. Empirical formula: C₁₃H₁₂O; molecular formula: C₁₃H₁₂O

3. Empirical formula: CH; molecular formula: C₈H₈

4. Empirical formula: C₂HCl; molecular formula: C₆H₃Cl₃

5. Empirical formula: Cl₄K₂Pt; molecular formula: Cl₄K₂Pt

6. Empirical formula: C₂H₄Cl; molecular formula: C₄H₈Cl₂

6 0

3 years ago

5,6-dimethyl-2-heptyne Due in 1 hr pls help me

mina [271]

Answer:

Please see the attached pictures.

Explanation:

☆ To ensure that each carbon has 4 bonds, fill the other bonds with Hs.

4 0

3 years ago