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3 years ago

The electron cloud model describes the __of electrons in an atom.

Chemistry

1 answer:

Alex Ar [27]3 years ago

4 0

Answer:

it's location

Explanation:

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What pH value can be assigned to acids and bases, respectively?

goldfiish [28.3K]

Acids have a pH less than 7 (pH < 7)
Bases have a pH more than 7 (pH > 7)
A pH of 7 would be neutral

Hope this helped!

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3 years ago

When the water inside a living cell freezes, the ice crystals damage the cell. The wood frog is a unique creature that can survi

professor190 [17]

But i think its B but i need answer too

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3 years ago

Round to four significant figures 0.00238866

mixas84 [53]

0.0024 Is it rounded to four significant figures

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3 years ago

A student obtains a mixture of the chlorides of two unknown metals, X and Z. The percent by mass of X and the percent by mass of

Aleksandr-060686 [28]

<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

<u>Explanation:</u>

To calculate the mole percent of a substance, we use the equation:

$\text{Mole percent of a substance}=\frac{\text{Moles of a substance}}{\text{Total moles}}\times 100$

Mass percent means that the mass of a substance is present in 100 grams of mixture

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively

Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

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3 years ago

Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907

melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent. $\frac{^0 C}{m}$

m = molality of solute (m or $\frac{mol}{Kg}$ )

Given : kf = 1.86 $\frac{^0 C*Kg}{mol}$

m = 0.907 $\frac{mol}{Kg}$ )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 $\frac{^0 C*Kg}{mol}$ * 0.907 $\frac{mol}{Kg}$ )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0

3 years ago