All categories

2 years ago

8. Which gas contributes the largest part of air? *

Chemistry

1 answer:

timurjin [86]2 years ago

5 0

Answer:

Nitrogen

Explanation:

Nitrogen present 78% in the earth's atmosphere

You might be interested in

2.In a titration, 25.00 cm3 of a solution of hydrochloric acid reacted with 18.40 cm3 of sodium hydroxide solution of concentrat

telo118 [61]

Answer:

0.11mol/dm³

Explanation:

The reaction expression is given as:

HCl + NaOH → NaCl + H₂O

Volume of acid = 25cm³ = 0.025dm³

Volume of base = 18.4cm³ = 0.0184dm³

Concentration of base = 0.15mol/dm³

Solution:

The concentration of hydrochloric acid = ?

To solve this problem, let us first find the number of moles of the base;

Number of moles = concentration x volume

Number of moles = 0.15mol/dm³ x 0.0184dm³ = 0.00276mol

From the balanced reaction equation;

1 mole of NaOH will combine with 1 mole of HCl

Therefore, 0.00276mol of the base will combine with 0.00276mol of HCl

So;

Concentration of acid = $\frac{number of moles }{volume}$ = $\frac{ 0.00276}{0.025}$ = 0.11mol/dm³

4 0

2 years ago

Which of the following is NOT true regarding Rutherford's Gold Foil experiment?

erastova [34]

Answer:

The area around the nucleus must be of low mass.

Explanation:

Rutherford`s experiment showed that there are some positive charges in the center of the atoms, and because they are all together, they will give a great mass to the atom.

It was quite different from Thomson`s experiment, in which it was thought that the negative charges were mixed with the positive charges, around the atom (like a Pudding Model). In Rutherford`s experiment, because the direction of beta particles, it was the prediction of the positive nucleus.

Hope this info is useful.

8 0

2 years ago

Plz help ASAP. thx for you answers ALGEBRA 2 sorry

iren [92.7K]

So, the answer to 27.) would be <em>2x.</em> Both 6x and 2x can be divided by 2x, but they can't go any higher without the end-answer becoming a fraction. As such, 2x is the greatest common factor.

For 28.), x and x^2 can't be like terms, since like terms have the same variable and exponent :)

Hope I could help!

3 0

3 years ago

The specific heats at constant pressure of some common gases are provided as a thirdorder polynomial: �;<<< = � + �� +

3241004551 [841]

Answer:

1.991 kJ

Explanation:

Calculate the amount of heat ( J )

CH4 ;

coefficients are : a = 19.89 , b = 5.02 * 10^-2 , c = 1.269 * 10^-5 , d = -11.01 * 10^-9

attached below is the detailed solution

3 0

2 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago