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2 years ago

Which of the following is an inner transition element?

Chemistry

1 answer:

KiRa [710]2 years ago

4 0

Option E) Th is correct answer. Thorium (Th) is inner transition element.

The components that make up the f-block are those where the 4f and 5f orbitals gradually fill up.
Although they are officially members of group 3, these elements are displayed separately in the f-block of the periodic table. Inner transition elements are another name for the f-constituent block's parts.
Lanthanoids and actinoids are the names for the two series of inner transition elements that are 4f and 5f series, respectively.
The inner transition elements U, Th, and Pa are excellent nuclear energy sources.
The inner transition elements' electronic configuration is 4f1-14 5d0-16 s².
Due to their uniform size and outer shell design, elements in a series would all have comparable qualities, making it difficult to distinguish them from one another.

Hence, thorium is inner transition element.

Learn more about lanthanoids here:

brainly.com/question/13760918

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For the following reaction, 9.60 grams of butane (C4H10) are allowed to react with 17.0 grams of oxygen gas. butane (C4H10) (g)

dangina [55]

Answer:

14.4 g of $CO_{2}$ can be produced.

Explanation:

Balanced equation: $2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O$

Molar mass (g/mol)

$C_{4}H_{10}$ 58.12

$O_{2}$ 32

$CO_{2}$ 44.01

So, 9.60 g of $C_{4}H_{10}$ = $\frac{9.60}{58.12}mol=0.165mol$

17.0 g of $O_{2}$ = $\frac{17.0}{32}mol=0.531mol$

According to balanced equation-

2 moles of $C_{4}H_{10}$ produce 8 moles of $CO_{2}$

So, 0.165 moles of $C_{4}H_{10}$ produce $(\frac{8}{2}\times 0.165)mol$ of $CO_{2}$ or 0.660 moles of $CO_{2}$

13 moles of $O_{2}$ produce 8 moles of $CO_{2}$

So, 0.531 moles of $O_{2}$ produce $(\frac{8}{13}\times 0.531)moles$ of $CO_{2}$ or 0.327 moles of $CO_{2}$

As least number of moles of $CO_{2}$ are produced from $O_{2}$ therefore $O_{2}$ is the limiting reagent.

So, maximum amount of $CO_{2}$ that can be formed = 0.327 moles

= $(0.327\times 44.01)g$

= 14.4 g

3 0

3 years ago

Draw the structural formula for 1 bromo 3 phenyl 5 heptyne

MrRissso [65]

See Figure 4 below.

1. Draw a 7-carbon chain and number the carbon atoms (Figure 1).

2. Insert a triple bond between atoms 5 and 6 (Figure 2).

3. Attach a phenyl group to atom 3 (Figure 3).

4. Attach a Br to atom 1 (Figure 4).

And you have your structure.

Note: The C≡C triple bond takes priority in the naming, so we must number the chain from the other end. The correct name is

7-bromo-5-phenylhept-2-yne (all one word).

6 0

3 years ago

The cell potential for an electrochemical cell with a Zn, Zn2 half-cell and an Al, Al3 half-cell is _____ V. Enter your answer t

kap26 [50]

Answer:

0.900 V

Explanation:

Oxidation half cell;

2Al(s) -----> 2Al^3+(aq) + 6e

Reduction half equation;

3Zn^2+(aq) + 6e ----> 3Zn(s)

E°anode = -1.66V

E°cathode= -0.76 V

E°cell= E°cathode - E°anode

E°cell= -0.76-(-1.66)

E°cell= 0.900 V

3 0

4 years ago

Calculate the molarity of 8.42 x 102 mL of solution containing 22.4 g of potassium iodide.

givi [52]

Answer:

Molarity = 0.154 M

Explanation:

Given data:

Volume of solution = 8.42 × 10² mL ( 8.42 × 10² /1000 = 0.842 L)

Mass of potassium iodide = 22.4 g

Molarity of solution = ?

Solution:

Number of moles of potassium iodide:

Number of moles = mass/molar mass

Number of moles = 22.4 g/ 166.0 g/mol

Number of moles = 0.13 mol

Molarity:

Molarity = number of moles / volume in L

Molarity = 0.13 mol / 0.842 L

Molarity = 0.154 M

8 0

3 years ago

Ann drove to the store km north of her house and then drove to the library, which is 5 km south of the store. She drove a total

Vadim26 [7]

Answer : The question is incomplete, the complete question is attached in answer

To calculate the displacement by Ann's driving from north of her house to south;

we have the data as

In north - 10 km
in south - 5 km
total - 15 km

So we can calculate the displacement, which will be,

10 - 5 = 5 km.

Hence, the displacement will be 5 km

3 0

3 years ago