What is activation energy?

20 points

9966 [12]

Answer:

I`m pretty sure it`s electrical tape.

Explanation:

8 0

2 years ago

Name two luminous objects other than the sun

jekas [21]

Answer: flame in a lamp, tube light, electric bulb

Explanation:

3 0

2 years ago

Read 2 more answers

Determine the [OH-], pH, and pOH of a solution with a [H+] of 0.090 M at 25 °C. [OH-] = pH = pOH = 0 POH =

ValentinkaMS [17]

Answer : The concentration of $OH^-$ ion, pH and pOH of solution is, $1.12\times 10^{-13}M$ , 1.05 and 12.95 respectively.

Explanation : Given,

Concentration of $H^+$ ion = 0.090 M

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

$pH=-\log [H^+]$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.090)$

$pH=1.05$

The pH of the solution is, 1.05

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-1.05=12.95$

The pOH of the solution is, 12.95

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12.95=-\log [OH^-]$

$[OH^-]=1.12\times 10^{-13}M$

The $OH^-$ concentration is, $1.12\times 10^{-13}M$

5 0

2 years ago

Read 2 more answers

What is the melting point of a solution in which 3.5 grams of sodium chloride is added to 230 mL of water?

kirza4 [7]

We are going to use this equation:

ΔT = - i m Kf

when m is the molality of a solution

i = 2

and ΔT is the change in melting point = T2- 0 °C

and Kf is cryoscopic constant = 1.86C/m

now we need to calculate the molality so we have to get the moles of NaCl first:

moles of NaCl = mass / molar mass

= 3.5 g / 58.44

= 0.0599 moles

when the density of water = 1 g / mL and the volume =230 L

∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg

now we can get the molality = moles NaCl / Kg water

=0.0599moles/0.23Kg

= 0.26 m

∴T2-0 = - 2 * 0.26 *1.86

∴T2 = -0.967 °C

8 0

3 years ago

Read 2 more answers

A gas has a volume of 4.25 m3 at a temperature of 95.0°C and a pressure of 1.05 atm. What temperature will the gas have at a pre

Goryan [66]

Answer:

$\boxed {\boxed {\sf 82.7 \textdegree C}}$

Explanation:

We are asked to find the temperature of a gas given a change in pressure and volume. We will use the Combined Gas Law, which combines 3 gas laws: Boyle's, Charles's, and Gay-Lussac's.

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

Initially, the gas has a pressure of 1.05 atmospheres, a volume of 4.25 cubic meters, and a temperature of 95.0 degrees Celsius.

$\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{P_2V_2}{T_2}$

Then, the pressure increases to 1.58 atmospheres and the volume decreases to 2.46 cubic meters.

$\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{1.58 \ atm *2.46 \ m^3}{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator by the second denominator, then multiply the first denominator by the second numerator.

$(1.05 \ atm * 4.25 \ m^3) * T_2 = (95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)$

Now the variable is being multiplied by (1.05 atm * 4.25 m³). The inverse operation of multiplication is division, so we divide both sides by this value.

$\frac {(1.05 \ atm * 4.25 \ m^3) * T_2}{(1.05 \ atm * 4.25 \ m^3)} = \frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}$

$T_2=\frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}$

The units of atmospheres and cubic meters cancel.

$T_2=\frac{(95.0 \textdegree C)*(1.58* 2.46 )}{(1.05 * 4.25 )}$

Solve inside the parentheses.

$T_2= \frac{(95.0 \textdegree C)*3.8868}{4.4625}$

$T_2= \frac{369.246}{4.4625} \textdegree C}$

$T_2 = 82.74420168 \textdegree C$

The original values of volume, temperature, and pressure all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 4 in the hundredth place to the right tells us to leave the 7 in the tenths place.

$T_2 \approx 82.7 \textdegree C$

The temperature is approximately <u>82.7 degrees Celsius.</u>

3 0

2 years ago