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Marina86 [1]
4 years ago
14

Which of the following statements best describes the energy transformation that occurs when a log burns? Mechanical energy chang

es to heat and light energy. Chemical energy changes to heat and light energy. Heat and light energy changes to chemical energy. Mechanical energy changes to chemical energy.
Chemistry
2 answers:
AnnyKZ [126]4 years ago
7 0

Answer: Chemical energy changes to light and heat energy.

Explanation:

The energy that is found in the wooden log is the chemical energy which is found in the bonds of carbon.

When the wooden log is burnt, the bonds are broken and the energy is converted in the form of light and heat. This is the conversion of chemical energy into heat energy and light energy.

This is also known as combustion in which a hydrocarbon is burnt to produce light and heat and carbon dioxide is released during the process.

horsena [70]4 years ago
3 0
The fire rearranges the molecules of a log while it is burning making it a chemical change.
Your answer is C

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Anon25 [30]
Chemical formula is H2O2
7 0
3 years ago
How many uL are present in 250 mL of H20? (1 uL = 10^-6 Liters)​
Vanyuwa [196]

Answer:

250000 μL

Explanation:

If         1 L = 1000 mL

Then  X L = 250 mL

X = (1 × 250) / 1000 = 0.25 L

Now we can calculate the number of microliters (μL) in 0.25 L:

if        1 μL = 10⁻⁶ L

then   X μL = 0.25 L

X = (1 × 0.25) / 10⁻⁶  =250000 μL

4 0
3 years ago
What controls the amount of product formed in a chemical reaction
melisa1 [442]

Answer: limiting reactant controls the amount of product formed in a chemical reaction.

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5 0
4 years ago
Compared to the freezing point of 1.0 M KCl(aq) at standard pressure, the freezing point of 1.0 M CaCl2(aq) at standard pressure
Galina-37 [17]

Answer : The correct option is, (1) lower

Explanation :

Formula used for lowering in freezing point :

\Delta T_f=i\times k_f\times m

where,

\DeltaT_b = change in freezing point

k_b = freezing point constant

m = molality

i = Van't Hoff factor

According to the question, we conclude that the molality of the given solutions are the same. So, the freezing point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) The dissociation of KCl will be,

KCl\rightarrow K^++Cl^-

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(b) The dissociation of CaCl_2 will be,

CaCl_2\rightarrow Ca^{2+}+2Cl^-

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The freezing point depends only on the Van't Hoff factor. That means higher the Van't Hoff factor, lower will be the freezing point and vice-versa.

Thus, compared to the freezing point of 1.0 M KCl(aq) at standard pressure, the freezing point of 1.0 M CaCl_2(aq) at standard pressure is lower.

4 0
3 years ago
A gas-filled balloon having a volume of 3.60 L at 1.15 atm and 20°C is allowed to rise to the stratosphere (about 30 km above th
MissTica

<u>Answer:</u> The new volume of the balloon will be 583.5 L

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

We are given:

P_1=1.15atm\\V_1=3.60L\\T_1=20^oC=[20+273]K=293K\\P_2=5.40\times 10^{-3}atm=0.00540atm\\V_2=?\\T_2=-50^oC=[-50+273]K=223K

Putting values in above equation, we get:

\frac{1.15atm\times 3.60L}{293K}=\frac{0.00540atm\times V_2}{223K}\\\\V_2=\frac{1.15\times 3.60\times 223}{293\times 0.00540}=583.5L

Hence, the new volume of the balloon will be 583.5 L

5 0
3 years ago
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