Ask question

Ask question

All categories

3 years ago

5

Hydrogen bonding occurs when hydrogen is bonded to N, O, or F. Which of the following molecules has hydrogen bonding, when bondi

ng with
their same molecules?
A. CBr4
O B.H2S
C. NO2
OD. NH3

1 answer:

Makovka662 [10]3 years ago

5 0

Answer:

I think the answer is……

O B.H2S

Explanation:

I’m not sure tho, I’m just not 100% positive.

You might be interested in

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 2.001 g sample of B

Alborosie

<u>Answer:</u> The empirical formula for the given compound is $C_{15}H_{24}O_1$

<u>Explanation:</u>

The chemical equation for the combustion of compound having carbon, hydrogen, iron and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=5.995g$

Mass of $H_2O=1.963g$

Mass of sample = 2.001 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 5.995 g of carbon dioxide, $\frac{12}{44}\times 5.995=1.635g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 1.963 g of water, $\frac{2}{18}\times 1.963=0.218g$ of hydrogen will be contained.

Mass of oxygen in the compound = (2.001) - (1.635 + 0.218) = 0.148 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.635g}{12g/mole}=0.136moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.218g}{1g/mole}=0.218moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.148g}{16g/mole}=0.0092moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0092 moles.

For Carbon = $\frac{0.136}{0.0092}=14.78\approx 15$

For Hydrogen = $\frac{0.218}{0.0092}=23.69\approx 24$

For Oxygen = $\frac{0.0092}{0.0092}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 15 : 24 : 1

Hence, the empirical formula for the given compound is $C_{15}H_{24}O_1$

3 0

4 years ago

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 1.80 fg (where a femtogram

egoroff_w [7]

Answer:

Δx ≥ 1.22 *10^-10m

Explanation:

<u>Step 1:</u> Data given

The E. coli bacterial cell has a mass of 1.80 fg ( = 1.80 * 10^-15 grams = 1.80 * 10^-18 kg)

Velocity of v = 8.00 μm/s (= 8.00 * 10^-6 m/s)

Uncertainty in the velocity = 3.00 %

E. coli bacterial cells are around 1 μm = 10^−6 m in length

<u>Step 2:</u> Calculate uncertainty in velocity

Δv = 0.03 * 8*10^-6 m/s =2.4 * 10^-7 m/s

<u>Step 3:</u> Calculate the uncertainty of the position of the bacterium

According to Heisenberg uncertainty principle,

Δx *Δp ≥ h/4π

Δx *mΔv ≥ h/4π

with Δx = TO BE DETERMINED

with m = 1.8 *10^-18 kg

with Δv = 2.4*10^-7

with h = constant of planck = 6.626 *10^-34

Δx ≥ 6.626*10^-34 / (4π*(1.8*10^-18)(2.4*10^-7))

Δx ≥ 1.22 *10^-10m

6 0

3 years ago

A 3.35 g sample of an unknown gas at 81 ∘C and 1.05 atm is stored in a 1.75 L flask.

ra1l [238]

Answer:

Molar mass = 52.96g/mol

density = 1.91g/L

Explanation:

using ideal gas equation

PV=nRT

7 0

4 years ago

Whats the answer to 7/2 × 1/5?

svp [43]

The answer to the problem is 7/10

3 0

3 years ago

Read 2 more answers

Calculate the frequency of visible light having a wavelength of 486 nm

hjlf

Ν= c/λ

λ= 486 nm * ( 1 m / 1x10^9 nm) = 4.86 x10^-7 m

v= (3.00 x 10^8 m/s) / (4.86 x10^-7 m)
v= 6.1728395x 10^14 s^-1
= 6.14 x10^-14 Hz

8 0

3 years ago