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vodomira [7]
3 years ago
6

Describe the energy transformation when a match burns.

Chemistry
1 answer:
devlian [24]3 years ago
3 0
The burning of a match is an exothermic reaction. Heat is being released as a result of the reaction.
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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
The half life of radon-222 is 3.8 days. How Much of a 100g sample is left after 15.2 days
professor190 [17]

Answer:  

6.2 g  

Explanation:  

In a first-order decay, the formula for the amount remaining after <em>n</em> half-lives is  

N = \frac{N_{0}}{2^{n}}  

where  

<em>N</em>₀ and <em>N</em> are the initial and final amounts of the substance  

1. Calculate the <em>number of half-lives</em>.  

If t_{\frac{1}{2}} = \text{3.8 da}  

n = \frac{t}{t_{\frac{1}{2}}} = \frac{\text{15.2 da}}{\text{3.8 da}}= \text{4.0}

2. Calculate the <em>final mass</em> of the substance.  

\text{N} = \frac{\text{100 g}}{2^{4.0}} = \frac{\text{100 g}}{16} = \text{6.2 g}

4 0
3 years ago
Grams of CuCI please help 10 point
rosijanka [135]
I DONOU I JEED HELP 629
5 0
3 years ago
Classify each of the following as either a physical change or a chemical change:
Taya2010 [7]

Answer:

physical change

a b c

chemical change

d

3 0
3 years ago
Read 2 more answers
Just as one dozen eggs always has 12 eggs in it, one mole of a
zzz [600]

Answer:

6.022x10^{23}atoms \ Al

Explanation:

Hello,

In this case, given the described concept regarding the Avogadro's number, we can easily notice that 27.0 g of aluminium foil has 6.022x10²³ atoms as shown below based on the mass-mole-particles relationship:

27.0gAl*\frac{1molAl}{27.0gAl} *\frac{6.022x10^{23}atoms \ Al}{1molAl} \\\\=6.022x10^{23}atoms \ Al

Notice this is backed up by the fact that aluminium molar mass if 27.0 g/mol.

Best regards.

8 0
3 years ago
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