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On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to t

Ne4ueva [31]

The given question is incomplete. The complete question is as follows.

On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to the appropriate blanks in the sentences on the right. make certain each sentence is complete before submitting your answer.

Sn, F, Ge, not predictable, Cr

Of Ge or Po, the larger atom is .........

Of F or Se, the larger atom is ..........

Of Sn or I, the larger atom is .........

Of Cr or W, the larger atom is ........

Explanation:

When we move down a group then there occurs an increase in atomic size of the atoms due to increase in the number of electrons.

Ge is a group 14 element which lies in period 4 and Po is a group 16 element that lies in period 6. As polonium is larger in size as compared to germanium.

Fluorine is a group 17 element and lies in period 2. Selenium is a group 16 element lies in 4. Therefore, selenium is larger in size as compared to fluorine.

Sn is a group 14 element that lies in period 5 and I is a group 17 element that lies in period 5. Hence, I is a larger atom.

Cr is a d-block element which lies in period 4 and W is also a d-block element which lies in period 6. Hence, W is larger in size than Cr.

Thus, we can conclude that given blanks are matched as follows.

Of Ge or Po, the larger atom is Po.

Of F or Se, the larger atom is Se.

Of Sn or I, the larger atom is I.

Of Cr or W, the larger atom is W.

7 0

3 years ago

Name a metal and non metal which are liquid at normal pressure and temperature<br>

liraira [26]

Answer:

metal : Mercury(Hg)

non metal : bromine (Br)

Explanation:

mercury is liquid at room temperature and pressure and the same as bromine

6 0

2 years ago

How many moles of CO are present in 35.88 L of the gas?

Pepsi [2]

Answer:

1.602 moles CO

Explanation:

To convert from liters to moles, divide by 22.4:

35.88 L / 22.4 = 1.602 moles CO

7 0

3 years ago

Which of the following statements is true about the composition of the atmosphere?

Phoenix [80]

Answer : Option D) The atmospheric conditions vary as one changes latitude and altitude.

Explanation : The composition of the atmosphere varies according to the latitude and altitude because of the unequal heating of the earth surface at different latitudes and altitudes which results into atmospheric changes. It also creates different regions and zones.

3 0

3 years ago

Read 2 more answers

Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.

weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0

3 years ago