What are elements that bond with actinium

Doping Se with P would produce a(n) ________ semiconductor with ____________ conductivity compared to pure Se.

deff fn [24]

Doping Se (group VI elements) with P(group V)elements would produce a P-TYPE semiconductor with HIGHER conductivity compared to pure Se

the reason is P dopant will introduce holes in the Se as P has lesser valence electron

4 0

3 years ago

Which of the following polyatomic ions will form an ionic compound with two sodium ions? CO32− HCO31− NO21− NO31−

BaLLatris [955]

Answer:

CO32−

Explanation:

We have to consider the valencies of the polyatomic ions involved. Recall that it is only a polyatomic ion with a valency of -2 that can form a compound which requires two sodium ions.

When we look closely at the options, we will realize that among all the options, only CO32− has a valency of -2, hence it must be the required answer. In order to be double sure, we put down the ionic reaction equation as follows;

2Na^+(aq) + CO3^2-(aq) ---------> Na2CO3(aq)

8 0

3 years ago

A sample of propane(c3h8)has 3.84x10^24 H atoms.

deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

$\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}$

= 4.75 × 10²³ molecules of Propane.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

= 14.25 × 10²³ 

________________________________________

Gram Molecular Mass of Propane(C3H8)

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

$\mathsf{ \frac{44 }{6.02 \times {10}^{23} } \times 4.75 \times {10}^{23} }$

$\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }$

$\mathsf{ = \frac{44 }{6.02 } \times 4.75 }$

= 34.72 g

8 0

2 years ago

One mole of acetyl chloride was mixed with one mole of dimethylamine. After the reaction is complete, what species can be found

fredd [130]

Answer:

N,N-dimethylacetamide is formed.

Explanation:

It is an example of a nucleophilic addition-elimination reaction. Here dimethylamine acts as a nucleophile.
In the first step, dimethyl amine gives nucleophilic addition reaction at carbonyl center of acetyl chloride.
In the second step, removal of Cl atoms occurs.
In the third step, deprotonation takes place from amino group to produce N,N-dimethylacetamide.
Full reaction mechanism has been shown below.

7 0

3 years ago

If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?

kodGreya [7K]

Answer:

$[Cu^{2+}]=0.041 M$

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

$[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}$

$[Cu^{2+}]=0.041 M$

Best regards!

6 0

3 years ago