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3 years ago

What is a field force that is used to generate electricity

Chemistry

1 answer:

miv72 [106K]3 years ago

6 0

A magnetic field is used to generate electricity

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explain why ionic compounds are not electrical conductors in the solid state, but conduct electricity when dissolved.

Alik [6]

Answer:

Hi !

Here is your answer !

Ionic compounds cannot conduct electricity when solid, as their ions are held in fixed positions and cannot move.

Explanation:

Ionic compounds conduct electricity when molten (liquid) or in aqueous solution (dissolved in water), because their ions are free to move from place to place.

Thank You !

4 0

3 years ago

How many hydrogen atoms are represented in the formula (CH3)2CH2?

SCORPION-xisa [38]

7 ..... i think because in CH3 there would be 3 H atoms while in 2CH2 there would be four

5 0

4 years ago

An amoeba is a one-celled organism. The cell theory states that which of the following characteristics of amoebas must be true?A

Yuki888 [10]

Answer:

I believe the answer is all amoeba come from other amoeba

6 0

4 years ago

Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

tekilochka [14]

Answer: There are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

So, the number of atoms present in 0.67 moles are as follows.

$0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms$

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

$4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms$

Thus, we can conclude that there are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

7 0

3 years ago

If 26.2 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity o

sweet-ann [11.9K]

Answer:

The molarity of the $AgNO_3$ solution is $4.02 \times 10^4 M$

Explanation:

The Balanced chemical equation is

$1AgNO_3 (aq) +1KCl (aq) > 1 AgCl (s)+1KNO_3 (aq)$

Mole ratio of $AgNO_3$ : KCl is 1 : 1

So moles $AgNO_3$ = moles KCl

$Moles KCl = \frac {mass}{molarmass}$

$= \frac {0.785 mg}{(39.1+35.5 g per mol)}$

$= \frac {0.000785 g}{74.6 g per mol}$

$= 0. 0000105 mol KCl$

$= 0.0000105 mol AgNO_3$

So Molarity

$= \frac {moles of solute}{(volume of solution in L)}$

$= \frac {0.0000105 mol}{26.2 mL}$

$=\frac {0.0000105 mol}{0.0262 L}$

= 0.000402M or mol/L is the Answer

(Or) $4.02 \times 10^4 M$ is the Answer

6 0

3 years ago