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Rudik [331]
3 years ago
6

What kind of solution would you have if it contained 50 grams of sodium chloride in 100 mL of water at 30˚C?

Chemistry
1 answer:
Svetlanka [38]3 years ago
5 0

Answer:

Supersaturated solution.

Explanation:

Hello!

In this case, according to the types of solution in terms of the relative amounts of solute and solvent, we can define a point called solubility at which the amount of solute is no longer dissolved in the solvent; thus, a value of solute/solvent less than the solubility is related to unsaturated solutions, equal to the solubility is related to the saturated solutions and more than the solubility to supersaturated solutions.

Thus, since solubility is temperature-dependent, at 30 °C the solubility of sodium chloride is 36.09 g per 100 mL of water; which means that, since the solution has 50 g of sodium chloride, more than 36.09 g, we infer this is a supersaturated solution.

Best regards!

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Read the given expression. X = number of protons − number of core electrons Which of the following explains the identity of X an
shepuryov [24]

<u>Answer: </u>The correct statement is X is the effective nuclear charge, and it increases across a period.

<u>Explanation:</u>

We are given that:

X = number of protons − number of core electrons

Effective nuclear charge is defined as the actual nuclear charge (Z = number of protons) minus the screening effect caused by the electrons present between nucleus and valence electrons. These electrons are the core electrons.

The formula used for the calculation of effective nuclear charge given by Slater is:

Z^*=Z-\sigma

where,

Z^* = effective nuclear charge

Z = atomic number or actual nuclear charge or number of protons

\sigma = Screening constant

The effective nuclear charge increases as we go from left to right in a period because nuclear charge increases with no effective increase in screening constant.

Hence, the correct answer is X is the effective nuclear charge, and it increases across a period.

5 0
3 years ago
How many molecules are contaiined in 125 grams of water, H2O?
vladimir1956 [14]

Answer:

Water has a molar mass of 18.015 g/mol . This means that one mole of water molecules has a mass of 18.015 g . So, to sum this up, 6.022⋅1023 molecules of water will amount to 1 mole of water, which in turn will have a mass of 18.015 g . 2.7144moles H2O ⋅6.022⋅1023molec.

Explanation:

7 0
2 years ago
How much heat is added if .0948g of water is increased in temperature by .728 degrees C?
Verdich [7]

Answer:

0.289J of heat are added

Explanation:

We can relate the change in heat of a substance with its increasing in temperature using the equation:

q = m*ΔT*S

<em>Where Q is change in heat</em>

<em>m is mass of substance (In this case, 0.0948g of water)</em>

<em>ΔT = 0.728°C</em>

<em>S is specific heat (For water, 4.184J/g°C)</em>

Replacing:

q = 0.0948g*0.728°C*4.184J/g°C

q = 0.289J of heat are added

5 0
3 years ago
What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?
PSYCHO15rus [73]

Answer:

V = 240.79 L

Explanation:

Given data:

Volume of butane = ?

Temperature = 293°C

Pressure = 10.934 Kpa

Mass of butane = 33.25 g

Solution:

Number of moles of butane:

Number of moles = mass/ molar mass

Number of moles = 33.25 g/ 58.12 g/mol

Number of mole s= 0.57 mol

Now we will convert the temperature and pressure units.

293 +273 = 566 K

Pressure = 10.934/101 = 0.11 atm

Volume of butane:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

V = nRT/P

V = 0.57 mol × 0.0821 atm.L/ mol.K  ×566 K  / 0.11 atm

V = 26.49 L/0.11

V = 240.79 L

6 0
3 years ago
If you were converting from milli- to centi- units, would you move the decimal point to the left or the right
yan [13]

Answer:

Left

Explanation:

6 0
3 years ago
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