The best electrolyte solutions are A. dilute B. concentrated C. saturated

24. The air in a small 250 cm birthday balloon is at a pressure of 760 torr. A boy sits on it at

kvv77 [185]

Answer:

1.14atm

Explanation:

Given parameters:

V1 = 250cm³ ;

1000cm³ = 1dm³; so this is 0.25dm³

P1 = 760torr

760torr = 1atm

V2 = 220cm³ ; 0.22dm³

Unknown:

New pressure = ?

Solution:

To solve this problem, we apply Boyle's law and we use the expression below:

P1 V1 = P2V2

The unknown is P2;

1 x 0.25 = P2 x 0.22

P2 = 1.14atm

7 0

3 years ago

1. a) Name two examples of good conductors?

Pachacha [2.7K]

A. good conductors - copper, aluminium

b. fair conductors - carbon,human body

c. insulator - paper, wood

7 0

4 years ago

what is the diffrence between displacement ans double displacement reactions write equations for these reactions

Lina20 [59]

Answer:

Single Displacement reaction

In a displacement reaction, a more reactive element replaces a less reactive element from a compound.

Change in colour takes place with no precipitate forms.

Metals react with the salt solution of another metal.

Examples:

2KI + Cl2 → 2KCl + I2

CuSO4 + Zn → ZnSO4 + Cu

Double displacement reaction

In a double displacement reaction, two atoms or a group of atoms switch places to form new compounds.

Precipitate is formed.

Salt solutions of two different metals react with each other.

Examples:

Na2SO4 + BaCl2 → BaSO4 + 2NaCl

2KBr + BaCl2 → 2KCl + BaBr2

Hope this helps...Please Mark as Brainliest!!

8 0

3 years ago

What is the molar mass of Ca(NO3)2?

Valentin [98]

The answer to the molar mass of Ca(NO3)2 is gonna be C. 164.1 g/mol

3 0

3 years ago

Read 2 more answers

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

4 years ago