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xz_007 [3.2K]
3 years ago
10

What is 5.2034 x 10^8 in standard format

Chemistry
2 answers:
777dan777 [17]3 years ago
8 0
Five hundred twenty million, three hundred and forty thousand. hope it helps!

Elanso [62]3 years ago
6 0
The answer is 4162720
hop this works than anyone else

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What is the simplest aromatic compound?<br> toluene<br> benzene<br> cyclohexane<br> cyclopropane
pav-90 [236]
It’s Benzene, I do believe :) hope this helped, good luck!
3 0
3 years ago
An isotope of iron has 28 neutrons. If the atomic mass of the isotope is 54, how many protons does it have?
Crazy boy [7]
26 because you need to do 54-28=26
The atomic mass number is the number of protons AND neutrons together.
5 0
3 years ago
Read 2 more answers
How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C
rjkz [21]

20600Cal              

Explanation:

Given parameters:

Mass of water = 319.5g

Initial temperature = 35.7°C

Final temperature = 100°C

Unknown:

Calories needed to heat the water = ?

Solution:

The calories is the amount of heat added to the water. This can be determined using;

     H  =   m  c Ф

c  = specific heat capacity of water = 4.186J/g°C

   H is the amount of heat

    Ф is the change in temperature

    H = m c (Ф₂ - Ф₁)

    H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J

Now;

     1kilocalorie = 4184J

     

85996.56J to kCal; \frac{85996.56}{4184}   = 20.6kCal  = 20600Cal

               

learn more:

Specific heat brainly.com/question/3032746

#learnwithBrainly

6 0
3 years ago
When 1.98g of a hydrocarbon is burned in a bomb calorimeter, the temperature increases by 2.06∘C. If the heat capacity of the ca
schepotkina [342]

Answer:

8.3 kJ

Explanation:

In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:

q for water:

q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g

                                      c: specific heat of water = 4.186 J/gºC

                                     ΔT : change in temperature = 2.06 ºC

so solving for q :

q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J

For calorimeter

q calorimeter  = C x  ΔT  where C: heat capacity of calorimeter = 69.6 ºC

                                     ΔT : change in temperature = 2.06 ºC

q calorimeter = 69.60J x 2.06 ºC = 143.4 J

Total heat released = 8,140 J +  143.4 J = 8,2836 J

Converting into kilojoules by dividing by 1000 we will have answered the question:

8,2836 J x 1 kJ/J = 8.3 kJ

7 0
3 years ago
A thermometer reads an outside air temperature of 35°c. What is the temperature in degrees Fahrenheit
Hoochie [10]

35°c is equal to 95°f

To do this multiply 35 and 1.8

35 x 1.8=63

Now add 32

Resulting in the answer 95

(The equation for to solve for c and f is c1.8+32=f

3 0
2 years ago
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