Give an example of an element, a molecule that has the element in it, and a mixture with the molecule.

blagie [28]

The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.

Option D

<h3>Chemical Reactions</h3>

Generally,the experiment shows that glass B temperature is higher than glass temperature A and this is given that observation that the solute dissolves faster in glass B than glass A.

Therefore,The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.

For more information on Temperature

brainly.com/question/13439286

5 0

2 years ago

Which term indicates how strongly an atom attracts the electrons in a chemical bond

Dmitrij [34]

Electronegativity is your answer.

6 0

2 years ago

Which set of numbers gives the correct possible values of I for n = 22

Alexandra [31]

Answer:

Hello friends

Explanation:

<h3>For a given principal quantum number for or n, the corresponding angular quantum number or is equivalent to a range between 0 and( n-1)</h3>

<h3>This means that the angular quantum number for a principal quantum number of 2 is equivalent to.</h3>

<h3>Hope it's helpfully. </h3>

4 0

2 years ago

A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is

Reika [66]

<u>Answer:</u> The pH of the buffer is 4.61

<u>Explanation:</u>

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of weak acid = 4.70

$[\text{conjuagate base}]}$ = moles of conjugate base = 3.25 moles

$[\text{acid}]$ = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

$pH=4.70+\log(\frac{3.25}{4.00})\\\\pH=4.61$

Hence, the pH of the buffer is 4.61

8 0

2 years ago

Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:

posledela

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of $HNO_3$ = $\frac{15.5g}{63g/mol\times 9.5L}=0.026M$

Equilibrium concentration of $NO$ = $\frac{16.6g}{30g/mol\times 9.5L}=0.058M$

Equilibrium concentration of $NO_2$ = $\frac{22.5g}{46g/mol\times 9.5L}=0.051M$

Equilibrium concentration of $H_2O$ = $\frac{189.0g}{18g/mol\times 9.5L}=1.10M$

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For the given chemical reaction:

$2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)$

The expression for $K_c$ is written as:

$K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}$

$K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}$

$K_c=3.72$

Thus the value of the equilibrium constant Kc for this reaction is 3.72

5 0

2 years ago