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bogdanovich [222]
1 year ago
15

Based on the solubility observations, which of the following pairs of cations could be distinguished by the addition of sodium c

hloride to the solutions?iron and calcium.
Chemistry
1 answer:
Licemer1 [7]1 year ago
7 0

Based on the solubility observations, barium & aluminum could be distinguished by the addition of sodium chloride to the solutions.

<h3>What happens when NaCl is added to a solution?</h3>
  • The ionic link that held sodium and chloride ions together is broken when water molecules force the ions apart.
  • The sodium and chloride atoms are encircled by water molecules after the salt compounds are separated. After that, the salt dissolves and forms a homogenous solution.
  • In order to keep patients from dehydrating, sodium chloride, an important nutrient, is employed in healthcare. It is employed as a spice to improve flavor and as a food preservative. Additionally, sodium chloride is employed in the production of polymers and other goods. Additionally, it is used to de-ice sidewalks and roadways.
  • Adding water to sodium chloride results in a physical change because no new product is created.

Learn more about sodium chloride added to a solution refer to :

brainly.com/question/28092739

#SPJ4

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What evidence would support Maria's claim that the fossilized remains are that of a bird, class Aves, instead of a turtle, class
lora16 [44]

Your answer would be C.

6 0
3 years ago
Read 2 more answers
The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that
kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is 1.1\times 10^6 M.

HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)

This value indicates that

A. CN^- is a stronger base than ONO^-

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is ONO^-

D. The conjugate acid of CN- is HCN

Answer: A. CN^- is a stronger base than ONO^-

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When K_{p}>1; the reaction is product favoured.

When K_{p}; ; the reaction is reactant favored.

When K_{p}=1; the reaction is in equilibrium.

As, K_p>>1, the reaction will be product favoured and as it is a acid base reaction where HONO acts as acid by donating H^+ ions and CN^- acts as base by accepting H^+

Thus HONO is a strong acid thus ONO^- will be a weak conjugate base and CN^- is a strong base which has weak HCN conjugate acid.

Thus the high value of K indicates that CN^- is a stronger base than ONO^-

7 0
3 years ago
A rigid vessel contains 3.98 kg of refrigerant-134a at 700 kPa and 60°C. Determine the volume of the vessel and the total intern
Arada [10]

<u>Answer:</u> The volume of the vessel is 0.1542m^3 and total internal energy is 162.0 kJ.

<u>Explanation:</u>

  • To calculate the volume of water, we use the equation given by ideal gas, which is:

PV=nRT

or,

PV=\frac{m}{M}RT

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g    (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

T = temperature of container = 60^oC=[60+273]K=333K

Putting values in above equation, we get:

700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L

Converting this value into m^3, we use the conversion factor:

1m^3=1000L

So, \Rightarrow (\frac{1m^3}{1000L})\times 154.21L

\Rightarrow 0.1542m^3

  • To calculate the internal energy, we use the equation:

U=\frac{3}{2}nRT

or,

U=\frac{3}{2}\frac{m}{M}RT

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g  (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = 8.314J/K.mol

T = temperature = 60^oC=[60+273]K=333K

Putting values in above equation, we get:

U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is 0.1542m^3 and total internal energy is 162.0 kJ.

6 0
3 years ago
Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring 12.0 ft×15.0 ft×8.60ft.
garik1379 [7]

<u>Given:</u>

Dimensions of the room= 12 ft * 15 ft * 8.60 ft

<u>To determine:</u>

The amount of HCN that gives the lethal dose in the room with the given dimensions

<u>Explanation:</u>

As per the World Health Organization, the lethal dose of HCN is around 300 ppm

300 ppm = 300 mg of HCN/ kg of inhaled air

Volume of air = volume of room = 12 * 15 *8.6 = 1548 ft³

Now,  1 ft³ = 28316.8 cm³

Therefore, the calculated volume of air corresponds to:

1548 * 28316.8 = 4.383 * 10⁷ cm3

Density of air (at room temperature 25 C) = 0.00118 g/cm3

Thus mass of air corresponding to the calculated volume is

Mass = Density * volume = 0.00118 g/cm3 * 4.383 * 10⁷ cm3

= 5.172*10⁴ g = 51.72 kg

Lethal amount of HCN corresponding to 51.72 kg of air would be.

= 51.72 kg air* 300 mg of HCN/1 kg air = 15516 mg

Ans: Lethal dose of HCN = 15.5 g

7 0
3 years ago
3. Would polar or non-polar gases be more likely to be considered an ideal gas? Explain
zubka84 [21]

Answer:

it is an ideal gas because  all of the substances that it has are good for us and won't affect us in any way

6 0
3 years ago
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