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3 years ago

15

when a person strikes a match, potential energy in the match is transformed into which types of energy?

2 answers:

Alinara [238K]3 years ago

6 0

Mechanical energy is energy associated with the motion and position of an object. When you strike a match, it moves through the air until it rubs against a surface. The rubbing produces the heat required to light the match. This is a transformation from mechanical energy to thermal (heat) energy.

natulia [17]3 years ago

4 0

Answer:

the answer is thermal and radiant

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Calculate the poh of this solution. round to the nearest hundredth. ph = 1.90 poh =

Viktor [21]

Answer:

The answer is 5.10

Explanation:

<h3><u>Given</u>;</h3>

pH = 1.9

<h3><u>To </u><u>Find</u>;</h3>

pOH = ?

We know that

pH + pOH = 7

pOH = 7 – pH

pOH = 7 – 1.90

pOH = 5.10

Thus, The pOH of the solution is 5.10

6 0

2 years ago

When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

3 0

2 years ago

What is the smallest particle of a covalent compound that has the properties of that compound?

Shtirlitz [24]

The smallest particle is the white blood cells because they are the ones that help you breathe

5 0

3 years ago

What is the correct prefix for 6?<br><br> O A. Oct-<br> O B. Hepta-<br> O C. Nano-<br> O D. Hexa-

Alchen [17]

Answer:

Hexa

Explanation:

Enjoy !

4 0

2 years ago

Read 2 more answers

Two systems with heat capacities 19.9 J mol-1 K-1 and 28.2 ] mol 1 K-1 respectively interact thermally and come to an equilibriu

MAVERICK [17]

Answer : The initial temperature of system 2 is, $19.415^oC$

Explanation :

In this problem we assumed that the total energy of the combined systems remains constant.

$-q_1=q_2$

$m\times c_1\times (T_f-T_1)=-m\times c_2\times (T_f-T_2)$

The mass remains same.

where,

$C_1$ = heat capacity of system 1 = 19.9 J/mole.K

$C_2$ = heat capacity of system 2 = 28.2 J/mole.K

$T_f$ = final temperature of system = $30^oC=273+30=303K$

$T_1$ = initial temperature of system 1 = $45^oC=273+45=318K$

$T_2$ = initial temperature of system 2 = ?

Now put all the given values in the above formula, we get

$-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K$

$T_2=292.415K$

$T_2=292.415-273=19.415^oC$

Therefore, the initial temperature of system 2 is, $19.415^oC$

8 0

3 years ago