Which type of noncovalent interaction can involve either the polypeptide backbone or amino acid side chains?

A 1. 07 g sample of a noble gas occupies a volume of 363 ml at 35°c and 678 mmhg. Identify the noble gas in this sample. (r = 0.

Margaret [11]

The identity of the noble gas is the sample is Krypton

<h3>Ideal Gas law</h3>

From the question, we are to determine the identity of the noble gas in the sample

From the ideal gas equation, we have that

PV = nRT

∴ n = PV / RT

Where P is the pressure

V is the volume

n is the number of moles

R is the gas constant

and T is the temperature

From the given information,

P = 678 mmHg = 0.892105 atm

V = 363 mL = 0.363 L

R = 0.08206 L.atm/mol.K

T = 35 °C = 35 + 273.15 K = 308.15 K

Putting the parameters into the equation, we get

n = (0.892105 × 0.363)/ (0.08206 × 308.15)

n = 0.0128 moles

Now, we will determine the Atomic mass of the sample

Using the formula,

Atomic = Mass / Number of moles

Atomic mass of the substance = 1.07 / 0.0128

Atomic mass of the substance = 83.6 amu

The noble gas with the closest atomic mass to this value is Krypton.

Molar mass of Krypton = 83.798 amu

Hence, the identity of the noble gas is the sample is Krypton

Learn more on Ideal Gas law here: brainly.com/question/20212888

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What statement best explains how life functions a unicellular organism are carried out?

s344n2d4d5 [400]

Answer:

The structures in the cell work together to perform its life functions

Explanation:

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3 years ago

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Drag each label to the correct location on the image identify the parts of the energy diagram

diamong [38]

If you would’ve attached a picture I’m sure it would’ve been a lot easier.

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Convert 125 g to kilograms

joja [24]

Move the decimal place to the left 3 digits.
0.125

3 0

3 years ago

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When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H

algol [13]

Answer: The empirical formula is $CH_2$ and molecular formula is $C_4H_8$

Explanation:

We are given:

Mass of $CO_2$ = 18.95 g

Mass of $H_2O$ = 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, = $\frac{12}{44}\times 18.59=5.07g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, = $\frac{2}{18}\times 7.759=0.862g$ of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.422}{0.422}=1$

For H = $\frac{0.862}{0.422}=2$

The ratio of C : H = 1: 2

Hence the empirical formula is $CH_2$ .

The empirical weight of $CH_2$ = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4$

The molecular formula will be= $4\times CH_2=C_4H_8$

6 0

3 years ago