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3 years ago

9

By what two characteristics are all the elements of the periodic table placed in a particular row and column

1 answer:

jarptica [38.1K]3 years ago

7 0

<span>Rows are defined as which shell of the element contains the outermost electrons. As the rows increase, the outermost shell value increases, as well. The columns are defined as which block contains those outermost electrons, and the shape at which that orbital takes.</span>

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2. What is the mass of one mole of Na?

borishaifa [10]

Sodium, Atomic mass: 22.989769 g
You can see in a periodic table

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3 years ago

How many atoms of carbon are in the sample

Serhud [2]

There are six atoms in the carbon

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2 years ago

What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)

lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

$\Delta T = i \times Kf \times m = 1 \times 6.90 \°C/m \times 1.324m = 9.14 \°C$

where,

i: van 't Hoff factor (1 for non-electrolytes)
Kf: cryoscopic constant
m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

$T = 80.26 \° C - 9.14 \° C = 71.12 \° C$

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

3 0

2 years ago

Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g

Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

Volume of base,

= 30 ml

Molarity of base,

= 0.05 m

Molar mass of acid,

= 400 g/mol

As we know,

⇒ $Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}$

On substituting the values, we get

⇒ $0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}$

⇒ $Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}$

⇒ $=1.5\times 10^{-3}$

hence,

⇒ $Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}$

On substituting the values, we get

⇒ $1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}$

⇒ $Mass \ of \ acid=1.5\times 10^{-3}\times 400$

⇒ $=0.60 \ g$

8 0

3 years ago

I Need help with this worksheet I Need the Work Please Thank you

Komok [63]

1. a. 3

b. 2

2. a. 4

b. 6

3. a. 1

b. 2

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2 years ago