What are the representative particles of elements, ionic compounds, and covalent compounds called?

When using a calorimeter, the initial temperature of a metal is 70.4C. The initial temperature of the water is 23.6C. At the end

Sunny_sXe [5.5K]

1) 29.8 C

At the beginning, the metal is at higher temperature (70.4 C) while the water is at lower temperature (23.6 C). When they are put in contact, the metal transfers heat to the water, until they reach thermal equilibrium: at thermal equilibrium the two objects (the metal and the water have same temperature). Therefore, since the temperature of the water at thermal equilibrium is 29.8 C, the final temperature of the metal must be the same (29.8 C).

2) 6.2 C

The temperature change of the water is given by the difference between its final temperature and its initial temperature:

$\Delta T = T_f - T_i$

where

$T_f = 29.8 C\\T_i = 23.6 C$

Substituting into the formula,

$\Delta T=29.8 C-23.6 C=6.2 C$

And the positive sign means that the temperature of the water has increased.

3) -40.6 C

The temperature change of the metal is given by the difference between its final temperature and its initial temperature:

$\Delta T = T_f - T_i$

where

$T_f = 29.8 C\\T_i = 70.4 C$

Substituting into the formula,

$\Delta T=29.8 C-70.4 C=-40.6 C$

And the negative sign means the temperature of the metal has decreased.

5 0

3 years ago

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onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re

solmaris [256]

Answer:

The volume of the final solution, V = 0.0305L

Explanation:

Number of moles = Concentration * volume

Concentration of HA = 1.00 * 10⁻⁴M

Volume of HA = 1000mL = 1 L

Number of moles of HA = 1.00 * 10⁻⁴ * 1

Number of moles of HA = 1.00 * 10⁻⁴ mols

Equation of reaction:

HA → H⁺ + A⁻

If 1 mol of HA produces 1 mol of H⁺ and A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of H⁺ and A⁻.

Since only 16% dissociation occurs = 0.16

Number of moles of H⁺ produced = 0.16 * 1.00 * 10⁻⁴

Number of moles of H⁺ produced = 1.6 * 10⁻⁵mols

Number of moles of A⁻ produced = 0.16 * 1.00 * 10⁻⁴

Number of moles of A⁻ produced = 1.6 * 10⁻⁵mols

Since 16% of HA dissociated into H⁺ and A⁻, 84% of HA is left

Number of mols of HA left = 0.84 * 1.00 * 10⁻⁴

Number of mols of HA left = 8.4 * 10⁻⁵mols

Concentration = num of moles/volume

Let the volume of the final solution be V

Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

Conc of A⁻ = 1.6 * 10⁻⁵/V

To calculate the dissociation constant

$k_{a} = [H^{+} ][A^{-} ]/[HA]$

$k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L$

3 0

3 years ago

Can I please have help

NeTakaya

Answer:

it will usually increase

Explanation:

potato

6 0

2 years ago

Two cylinders A&B at the same temperature contains the same quantity of the same kind of gas. Cylinder A has three times the

GREYUIT [131]

Answer:

pressure in cylinder A must be one third of pressure in cylinder B

Explanation:

We are told that the temperature and quantity of the gases in the 2 cylinders are same.

Thus, number of moles and temperature will be the same for both cylinders.

To this effect we will use the formula for ideal gas equation which is;

PV = nRT

Where;

P is prrssure

V is volume

n is number of moles

T is temperature

R is gas constant

We are told that Cylinder A has three times the volume of cylinder .

Thus;

V_a = 3V_b

For cylinder A;

Pressure = P_a

Volume = 3V_b

Number of moles = n

Thus;

P_a × 3V_b = nRT

For cylinder B;

Pressure = P_b

Volume = V_b

Number of moles = n

Thus,

P_b × V_b = nRT

Combining the equations for both cylinders, we have;

P_a × 3V_b = P_b × V_b

V_b will cancel out to give;

3P_a = P_b

Divide both sides by 3 to get;

P_a = ⅓P_b

Thus, pressure in cylinder A must be one third of pressure in cylinder B

3 0

3 years ago

Esters are commonly used as artificial ___

Elza [17]

Answer:

Fragrances

They are commonly used as fragrances and found in essential oils and pheromones.

4 0

3 years ago

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