Which of the following is what an elements are made of?

A LA TEMPERATURA DE 35°C, UNA MUESTRA DE DIOXIDO DE CARBONO OCUPA UN VOLUMEN DE 350 ML. ¿Qué CAMBIO DE VOLUMEN SE PRODUCIRA SI L

Kisachek [45]

Answer:

New volume = 150 mL

Explanation:

Initial temperature, T₁ = 35°C

Initial volume, V₁ = 350 mL

We need to find the change in volume when the temperature drops to 15°C.

The relation between the temperature and the volume is given by Charle's law. Let new volume is V₂. It can be given by :

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{350\times 15}{35}\\\\V_2=150\ mL$

So, the new volume is 150 mL.

8 0

3 years ago

onsider the reversible dissolution of lead(II) chloride. P b C l 2 ( s ) − ⇀ ↽ − P b 2 + ( a q ) + 2 C l − ( a q ) PbClX2(s)↽−−⇀

Sveta_85 [38]

Answer:

9.34x10^-4

Explanation:

Step 1:

The balanced equation for the reaction.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

Step 2:

Data obtained from the question:

Mass of PbCl2 = 0.2393 g

Volume = 50mL

concentration of Pb^2+, [Pb^2+] = 0.0159 M

Concentration of Cl^-, [Cl^-] = 0.0318 M

Equilibrium constant, Kc =?

Step 3:

Determination of the number of mole PbCl2.

The number of mole of PbCl2 can be obtained as follow:

Molar Mass of PbCl2 = 207 + (35.5x2) = 278g/mol

Mass of PbCl2 = 0.2393 g

Number of mole =Mass /Molar Mass

Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

Step 4:

Determination of Molarity of PbCl2.

At this stage we shall obtain the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.61x10^-4 mole

Volume = 50mL = 50/1000 = 0.05L

Molarity of PbCl2 =?

Molarity = mole /Volume

Molarity of PbCl2 = 8.61x10^-4/0.05

Molarity of PbCl2 = 0.01722 M

Step 5:

Determination of the equilibrium constant Kc.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

The equilibrium constant Kc for the equation above is given by:

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

[Pb^2+] = 0.0159 M

[Cl^-] = 0.0318 M

[PbCl2] = 0.01722 M

Kc =?

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2/ 0.01722

Kc = 9.34x10^-4

5 0

3 years ago

A lead ball is added to a graduated cylinder containing 50.6 ml of water, causing the level of the water to increase to 93.0 mL.

Kamila [148]

42.4 ml is the volume in milliliters of the lead ball if a lead ball is added to a graduated cylinder containing 50.6 ml of water.

<h3>What is a graduated cylinder?</h3>

A tall narrow container with a volume scale is used especially for measuring liquids.

The graduated cylinder contains water

mL is a volume unit.

Water volume = 50.6 ml

The lead ball caused an increase in volume from 50.6 ml to 93.0 mL.

The new volume is the lead ball volume plus the original water volume :

Final volume = Vlead ball+ Water original volume

$93.0 mL= V_(lead ball) +50.6 ml$

$V_(lead ball) = 93.0 mL - 50.6 ml$

$V_(lead ball) = 42.4 ml$

Hence, 42.4 ml is the volume in milliliters of the lead ball.

Learn more about the graduated cylinder here:

brainly.com/question/13386106

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4 0

2 years ago

Record the volume of the liquid in the graduated cylinder. The volume of the liquid is

mario62 [17]

Answer:

53

Explanation:

5 0

3 years ago

Hydrogen (H) and Beryllium (Be) are members of the same <br> a Group <br> b Period <br> c Neither

Solnce55 [7]

Answer:

C. Neither

Explanation:

According to the periodic table, Hydrogen is in group 1 and period 1. Beryllium is in group 2, period 2.

3 0

3 years ago

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