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andrezito [222]
3 years ago
11

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]

<u>Answer:</u>

<em>The molarity of the AgNO_3 solution is 4.02 \times 10^4 M </em>

<em></em>

<u>Explanation:</u>

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
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