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2 years ago

Which phrase describes valence electrons? (1 point)

Chemistry

2 answers:

dimaraw [331]2 years ago

7 0

Answer:

the electrons in an atom's outermost energy level

Explanation:

Anna11 [10]2 years ago

4 0

Answer:

Explanation:

the electrons in an atom's outermost energy level

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If the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in

anygoal [31]

Answer:

d. supersaturated.

Explanation:

A solution naturally contains a solute and a solvent. The solute is the solid substance that dissolves in the solvent, which is usually a liquid substance. A solution has a maximum amount of solute that can dissolve in its constituent solvent.

However, when the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in the solution at that temperature, the solution is said to be SUPERSATURATED. This means that the solution contains more than the maximum amount of solute.

5 0

2 years ago

Which of the following leads to a higher rate of diffusion?

miv72 [106K]

<h3><u>Answer;</u></h3>

Higher velocity of particles

<h3><u>Explanation;</u></h3>

The diffusion rate is determined by a variety of factors which includes;

Temperature such that the higher the temperature, the more kinetic energy the particles will have, so they will move and mix more quickly and the diffusion rate will be high.
Concentration gradient such that the greater the difference in concentration, the quicker the rate of diffusion.
Higher velocity of particles increases the diffusion rate as this means more kinetic energy by the particles and hence the particles will mix and move faster, thus higher diffusion rate.

5 0

2 years ago

Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. T

Ostrovityanka [42]

Explanation:

The given data is as follows.

$\Delta H$ = 286 kJ = $286 kJ \times \frac{1000 J}{1 kJ}$

= 286000 J

$S_{H_{2}O} = 70 J/^{o}K$ , $S_{H_{2}} = 131 J/^{o}K$

$S_{O_{2}} = 205 J/^{o}K$

Hence, formula to calculate entropy change of the reaction is as follows.

$\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)$

= $[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]$

= $[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]$

= 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

$\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}$

= $286000 J - (163.5 J/K \times 298 K)$

= 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

6 0

3 years ago

What is the final volume of NaOH solution prepared from 100.0 mL of 0.500 M NaOH if you wanted the final concentration to be 0.1

suter [353]

Answer:

The final volume of NaOH solution is 30ml

Explanation:

We all know that

V1S1 = V2S2

or V1= V2S2÷S1

or V1= V2×S2×1/S1

or V1=100×0.15×1/0.50

V1= 30

∴30 ml NaOH solution is required to prepare 0.15 M from 100ml 0.50 M NaOH solution.

5 0

2 years ago

What volume of carbon dioxide, at 1 atm pressure and 112°C, will be produced when 80.0 grams of methane is burned?

Vadim26 [7]

Answer:

158 L.

Explanation:

What is given?

Pressure (P) = 1 atm.

Temperature (T) = 112 °C + 273 = 385 K.

Mass of methane CH4 (g) = 80.0 g.

Molar mass of methane CH4 = 16 g/mol.

R constant = 0.0821 L*atm/mol*K.

What do we need? Volume (V).

Step-by-step solution:

To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:

$PV=nRT.$

Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.

So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:

$80.0g\text{ CH}_4\cdot\frac{1\text{ mol CH}_4}{16\text{ g CH}_4}=5\text{ moles CH}_4.$

So, in this case, n=5.

Now, let's solve for 'V' and replace the given values in the ideal gas law equation:

$V=\frac{nRT}{P}=\frac{5\text{ moles }\cdot0.0821\frac{L\cdot atm}{mol\cdot K}\cdot385K}{1\text{ atm}}=158.04\text{ L}\approx158\text{ L.}$

The volume would be 158 L.

6 0

6 months ago