All categories

1 year ago

Rank the following salts in order of increasing pH of their 0.1 M aqueous solutions:NH₄NO₃ , NaHSO₄ , NaHCO₃ , Na₂CO₃

Chemistry

1 answer:

yan [13]1 year ago

3 0

Salts in order of growing pH of their 0.1 M aqueous solutions: NaHCO₃, , NaHSO₄ , Na2CO3, NH₄NO₃ .

<h3>How do you determine the pH of a substance?</h3>

To calculate the pH of an aqueous solution you want to recognize the attention of the hydronium ion in moles per liter (molarity). The pH is then calculated using the expression: pH = - log [H3O+].

<h3>What makes a salt acidic?</h3>

Since the cation acts as weak, a salt with an anion of a strong acid and a cation of a susceptible base produces an acidic solution with a pH much less than 7.

In this case, the anion will become the spectator ion, while the weak acid dissociates in water and yields H3O+ ions.

Learn more about pH of salts here:

<h3>brainly.com/question/1352821</h3><h3 /><h3>#SPJ4</h3>

You might be interested in

To begin the experiment, 1.11g of methane CH4is burned in a bomb calorimeter containing 1000 grams of water. The initial tempera

navik [9.2K]

1. CH4 + 2O2 → CO2 + 2H2O
2. 10.80 degrees celsius
3. 45,187 J
4. 7,506 J
<span>5. 52,693 J </span>

6 0

2 years ago

A piece of metal with a mass of 611 g is placed into a graduated cylinder that contains 25.1 ml of water, raising the water leve

jeka94

Mass of metal piece is 611 g and volume of graduated cylinder is 25.1 mL. When metal piece is placed in the graduated cylinder water level increases to 56.7 mL. The increase in volume is due to volume of metal piece that gets added to the volume of water.

Thus, volume of metal piece can be calculated by subtracting initial volume from the final one.

$V_{metal}=V_{final}-V_{initial}=(56.7-25.1)mL=31.6 mL$

Thus, volume of metal piece will be 31.6 mL. The mass of metal piece is given 611 g, density of metal can be calculated as follows:

$d=\frac{m}{V}=\frac{611 g}{31.6 mL}=19.33 g/mL$

Therefore, density of metal is 19.33 g/mL.

3 0

3 years ago

what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?

IgorLugansk [536]

Answer: 12.78ml

Explanation:

Given that:

Volume of KOH Vb = ?

Concentration of KOH Cb = 0.149 m

Volume of HBr Va = 17.0 ml

Concentration of HBr Ca = 0.112 m

The equation is as follows

HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)

and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb

(0.112 x 17.0)/(0.149 x Vb) = 1/1

(1.904)/(0.149Vb) = 1/1

cross multiply

1.904 x 1 = 0.149Vb x 1

1.904 = 0.149Vb

divide both sides by 0.149

1.904/0.149 = 0.149Vb/0.149

12.78ml = Vb

Thus, 12.78 ml of potassium hydroxide solution is required.

5 0

2 years ago

I have no idea what to put, pls help, due tmr<br>(science)

rodikova [14]

Answer:

It traveled for 4 seconds at a distance of about 4.1 meters. With a velocity of 10%.

Explanation:

6 0

2 years ago

By mistake, a quart of oil was dumped into a swimming pool that measures 25.0 m by 30.0 m. The density of the oil was 0.750 g/cm

kondor19780726 [428]

The oil slick thick = 1.256 x 10⁻⁴ cm

<h3>Further explanation</h3>

Volume is a derivative quantity derived from the length of the principal

The unit of volume can be expressed in liters or milliliters or cubic meters

The conversion is

1 cc = 1 cm3

1 dm = 1 Liter

1 L = 1.06 quart

<em>so for 1 quart = 0.943 L</em>

$\tt 0.943~L=0.943\times 10^{-3}m^3$

Volume of oil dumped = volume of swimming pool

$\tt 0.943\times 10^{-3}~m^3=25\times 30\times h(h=thick)\\\\h=\dfrac{0.943\times 10^{-3}}{750~m^2}=1.257\times 10^{-6}~m=\boxed{\bold{1.256\times 10^{-4}~cm}}$

3 0

3 years ago