All categories

2 years ago

When you strike a match, what do you provide that enables the flame to appear?

Chemistry

1 answer:

Alexeev081 [22]2 years ago

3 0

Explanation:

The minimum amount of energy required by a substance to begin a reaction is known as the activation energy.

Therefore, we can conclude that when you strike a match, you provide activation energy that enables the flame to appear.

Whereas a catalyst helps in increasing the rate of reaction but it does not start the reaction. On the other hand, when reactants combine together then we get products.

Thus, we can conclude that options (a), (b), (c), and (e) are not correct.

Hence, option (d) is the correct answer.

You might be interested in

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

lions [1.4K]

Answer: The equilibrium concentration of $COF_2$ is 0.332 M

Explanation:

We are given:

Initial concentration of $COF_2$ = 2.00 M

The given chemical equation follows:

$2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)$

Initial: 2.00

At eqllm: 2.00-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}$

We are given:

$K_c=6.30$

Putting values in above expression, we get:

$6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25$

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of $COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M$

Hence, the equilibrium concentration of $COF_2$ is 0.332 M

4 0

3 years ago

. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r

Tom [10]

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

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

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

3 0

3 years ago

What is the mass of a bacterial cell?

SVEN [57.7K]

Thus, it was found that the intestinal microbiota has approximately 100 trillion bacterial cells - about ten times more than the total number of cells in the human body.

6 0

2 years ago

Read 2 more answers

A small motor does 4000 J of work in 20 s. What is the power of the motor in watts?

TEA [102]

Answer:

200 W

Explanation:

Formula of Power is

P= W/T

where

W= Work and T= Time

4000/20 = 200J/s or 200 W

6 0

2 years ago

Calculate the volume of 4.50 mol of SO2 (g) measured at STP. At STP: P = 101.3 kPa, T = 273.15 K A. 87.3 L B. 101 L C. 127 L D.

miss Akunina [59]

Answer:

101L

Explanation:

We use the ideal gas relation

PV = nRT

P = pressure = 101.3KPa

V = volume = ?

n = number of moles = 4.5moles

T = Temperature = 273.15K

R = molar gas constant = 8.314J/mol.k

Rearranging the equation to make V the subject of the formula yields :

V = nRT/P

= ( 4.5 × 8.314 × 273.15) ÷ 101.3

= 10,219.361 ÷ 101.3 = 100.88L which is apprx 101L

8 0

3 years ago