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Leviafan [203]
3 years ago
8

A new stable element with an atomic number of 120 and an atomic mass of 246 is created in a particle accelerator. Enough of this

element is created to determine that it combines with chlorine in a 1:2 ratio. Element 120 should be placed in the same group as _____ on the periodic table.
Chemistry
1 answer:
deff fn [24]3 years ago
3 0

Answer:

Be,Mg,Ra etc

Explanation:

It should be palced in group 2A because as it reacts with chlorine in ratio of 1:2 . It's valancy is 2 and is metal as it react with non metal donating two electrons .

one more thing it fits there orderly

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If 31.6 g of KMnO4 is dissolved in enough water to give 160 mL of solution, what is the molarity?
Zina [86]

Answer:

A. 1.25M

B. 19.98g

Explanation:

A. Data obtained from the question include the following:

Mass of KMnO4 = 31.6 g

Volume = 160 mL

Molarity =..?

We'll begin by calculating the number of mole KMnO4 in the solution. This is can be obtained as follow:

Mass of KMnO4 = 31.6 g

Molar mass of KMnO4 = 39 + 55 + (16x4) = 158g/mol

Number of mole of KMnO4 =..?

Mole = mass /Molar mass

Number of mole of KMnO4 = 31.6/158 = 0.2 mole

Now, we can obtain the molarity of the solution as follow:

Volume = 160 mL = 160/1000 = 0.16L

Mole of KMnO4 = 0.2 mole

Molarity = mole /Volume

Molarity = 0.2/0.16 = 1.25M

B. Data obtained from the question include the following:

Volume = 300mL

Molarity = 0.74 M

Mass of H2C2O4 =..?

First, we shall determine the number of mole H2C2O4. This is illustrated below:

Volume = 300mL = 300/1000 = 0.3L

Molarity = 0.74 M

Mole of H2C2O4 =?

Mole = Molarity x Volume

Mole of H2C2O4 = 0.74 x 0.3

Mole of H2C2O4 = 0.222 mole

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Number of mole of H2C2O4 = 0.222 mole

Molar mass of H2C2O4 = (2x1) + (12x2) + (16x4) = 2 + 24 + 64 = 90g/mol

Mass of H2C2O4 =..?

Mass = mole x molar mass

Mass of H2C2O4 = 0.222 x 90

Mass of H2C2O4 = 19.98g

5 0
3 years ago
What is a similarity between the ovule and the stigma?
vekshin1
It’s the first one they both produce the plants pollen
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3 years ago
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What is the pressure of 0.33 moles of nitrogen gas, if its volume is 15.0 L at –25.0oC?
musickatia [10]

Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=

V

nRT

​

 . Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}

mol⋅K

atm⋅L

​

 :

P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=

V

nRT

​

=

15.0

L

​

(0.33

mol

)(0.0821

mol⋅K

atm⋅

L

​

​

)(248.15

K

​

)

​

=0.4482atm

In conclusion, the pressure of this gas is P=0.4482 atm.

Reference:

Chang, R. (2010). Chemistry. McGraw-Hill, New York.

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Is carbon monoxide or oxygen more ideal gas
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Answer: Oxygen

Explanation:

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