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3 years ago

As you increase the temperature of a solute it's solubility

1 answer:

7 0

There would be a direct result as an increase in the solute temperature will result in increase in its solubility. A greater amount of solute molecules will possess more kinetic energy and will be distributed and in container.

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For elements (Br, Ca, Fe, Na, S, Si, and Xe) which of the following term(s) apply?

ioda

Answer:

Detail is given below.

Explanation:

Bromine: Br (Halogen and non metal)

Bromine is present in group 17 and it is called halogen.

All halogens are very reactive.

Calcium; Ca ( alkaline earth metal)

Calcium is present in group 2. It is alkaline earth metal.

It has two valance electrons.

Iron; Fe (Transition metal)

Iron is transition metal and also called d-block element.

Sodium: Na (Metal, main group element)

Sodium is metal and present in group one.

It has one valance electron.

Sulfur: S (non metal)

Sulfur is non metal. It is present in group 16.

It has six valance electrons.

Silicon: Si (metalloid)

Silicon is metalloid. It is present in group 14.

It has four valance electrons.

Xenon: Xe (Noble gas, gas at a room temperature)

it is noble gas. Xenon is present in group 18.

It is gas at a room temperature.

3 0

3 years ago

Which radiation can only be stopped by thick layers of dense metal? (2 points)

AfilCa [17]

Gamma rays have the highest penetrating power so it can only be stopped by thick layers of dense metal. :)

5 0

3 years ago

Read 2 more answers

Determine the direction of bond polarity for each compound. Leave it blank for a nonpolar bond. Br−Br H−Cl F−CH3

NeX [460]

Answer:

For H-Cl, the direction is towards the chlorine atom

For F-CH3, the direction is towards the flourine atom.

Explanation:

The dipole moment is a vector quantity. This implies that it has both magnitude and direction.

Thus, the direction of the dipole moment always points from the positive atom towards the negative atom.

This explains the fact that it points to chlorine in HCl and points to flourine in F-CH3

8 0

3 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago

HELP QUICK ASSESMENT

ehidna [41]

Answer:

MAGNESIUM

Explanation:

7 0

2 years ago

Read 2 more answers

As you increase the temperature of a solute it's solubility​

As you increase the temperature of a solute it's solubility