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Otrada [13]
3 years ago
8

Which electromagnetic waves are used in ovens and cell phone communications

Chemistry
2 answers:
yuradex [85]3 years ago
8 0
Thermal energy
hope it helps!
densk [106]3 years ago
7 0
Ovens use microwaves. Believe it or not, thats the name
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H 2 O, CO 2 , and C 12 H 22 O 11 are all examples of chemical _____.
Reil [10]
They are all examples of chemical formulas
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2 years ago
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What would be oil classification?
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The service rating of passengers car and commercial automotive motor oils
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3 years ago
How many moles of liquid water must freeze to remove 100 kJ of heat? (ΔHf = –334 J/<br> g.
Juliette [100K]
First figure out how many grams must freeze and then convert the grams to moles. 
<span>Hf = -334 J/g. Convert this to KJ/g by dividing by 1000. (There are 1000 Joules in a kJ). </span>
<span>Hf = -334 J/g ÷ 1000 J/kj = -0.334 kJ/g </span>
<span>Now, divide 100 kJ by -0.334 kJ/g (see how the units are lining up?) </span>
<span>100 kJ ÷ -0.334 kJ/g = 299 g </span>
<span>Now convert this to moles by dividing by the molecular weight of water (18.0g/mole). </span>
<span>299 ÷ 18.0 = 16.6 moles </span>
5 0
3 years ago
What volume does 0.482 mol of gas occupy at a pressure of 719 mmHg at 56 ∘C?
kozerog [31]

Answer:

The volume is 13, 69 L

Explanation:

We use the formula PV=nRT. We convert the temperature in Celsius into Kelvin and the pressure in mmHg into atm.

0°C= 273K---> 56°C= 56 + 273= 329K

760 mmHg----1 atm

719 mmHg----x= (719 mmHgx 1 atm)/760 mmHg= 0,95 atm

PV=nRT ---> V= (nRT)/P

V=( 0,482 molx 0,082 l atm/K mol x 329K)/0,95 atm

<em>V=13,68778526 L</em>

7 0
3 years ago
Hi, can someone help me balance this chemistry equation:<br> H2SO4 + RbOH -&gt; Rb2SO4 + H2O
zloy xaker [14]

H2SO4 + 2RbOH -> Rb2SO4 + 2H2O

If you want an explanation, keep reading.

In the first portion, there are two hydrogen ions and four sulfate ions.

The second portion has one rubidium ions and one hydroxide ion.

On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.

Back to the right side, there is there is water (H2O).

On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).

So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).

I hope this was easy to understand.

6 0
3 years ago
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