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3 years ago

describe the differences of the kinetic energy of a cold water solution to that of a hot water solution

Chemistry

2 answers:

hjlf3 years ago

8 0

Answer:

Here are the differences between the hot and cold water. The particles in the hot water move faster than the particles in the cold water.

Explanation: I hope this helped you!!!!!!!:)

Natasha_Volkova [10]3 years ago

4 0

The kinetic energy is in relation to the temperature. In cold water, the molecules in the water moves slower and gives off less heat. Which also means it has a lower temperature and lower kinetic energy. Hot water is the opposite. The molecules move faster and have a higher kinetic energy. It has a higher temperature too.

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A hypothetical element consists of 2 isotopes of masses 84.95 amu and 86.95 amu with abundance of 37.1% and 62.9% respectively.

vodka [1.7K]

The average atomic mass of the hypothetical element is the sum of the products of the isotopes and their percentage abundance. For this given,
atm = (84.95 amu)(0.371) + (86.95 amu)(0.629)
= 86.208 amu
Thus, the average atomic mass of the element is 86.208 amu.

6 0

3 years ago

How much 3m stock solution would it take to make 70 ml of 1.2 m solution?

nordsb [41]

Answer:

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2 years ago

Green light has a wavelength of 550nm. What is the frequency of a photon of green light?

Vikki [24]

Answer:- Frequency is $5.4*10^1^4Hz$ .

Solution:- frequency and wavelength are inversely proportional to each other and the equation used is:

$\nu =\frac{c}{\lambda }$

where, $\nu$ is frequency, c is speed of light and $\lambda$ is the wavelength.

Speed of light is $3.0*10^8m/s$ .

We need to convert the wavelength from nm to m.

( $1nm=10^-^9m$ )

$550nm(\frac{10^-^9m}{1nm})$

= $5.5*10^-^7m$

Now, let's plug in the values in the equation to calculate the frequency:

$\nu =\frac{3.0*10^8m.s^-^1}{5.5*10^-^7m}$

= $5.4*10^1^4s^-^1$ or $5.4*10^1^4Hz$

since, $1s^-^1=1Hz$

So, the frequency of the green light photon is $5.4*10^1^4Hz$ .

4 0

3 years ago

On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed

KIM [24]

Answer:

On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?

A) BrCH2CH2Br

B) CH3CH2CH2Br

C) CH3CHBr2

D) CH3CH2CH2CH3

E) BrCH2CH2CH2CH2Br

Explanation:

The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.

This is an example of free radical substitution.

The structure of ethane and its bromination is shown below:

Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).

Remaining all other products are possisble to form on free radical substitution of ethane.

8 0

3 years ago

Use bond energies to calculate ΔHrxn Δ H r x n for the reaction. 2H2(g)+O2(g)→2H2O(g) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g )

Olenka [21]

Answer:

$\large \boxed{\text{-486 kJ}}$

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

2H₂ + O₂ ⟶ 2H-O-H

Bonds: 2H-H 1O=O 4H-O

D/kJ·mol⁻¹: 436 498 464

$\begin{array}{rcl}\Delta H & = & \sum{mD_{\text{reactants}}} - \sum{nD_{\text{products}}}\\& = & 2 \times 436 +1 \times 498 - 4 \times 464\\&=& 1370 - 1856\\&=&\textbf{-486 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{-486 kJ}}$}.$

3 0

3 years ago