Substances which will glow in the dark after being exposed to sunlight are:

matrenka [14]

Answer:

Yes.

Explanation:

Yes, this difference of readings will definitely affect the results of the experiment as well as the E values because the readings taken by both students are different from one another. There is a fault in one of the thermometer because both shows different readings of temperature of the same solution. This will affect the overall experiment and due to this error, we are unable to tell that which one reading is correct so the answer is uncertain or unsure.

5 0

2 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

3 years ago

The atoms, molecules, or compounds present at the start of a chemical reaction that participate in the reaction.

Art [367]

For the first one the answer is B. and the second one is D.

4 0

3 years ago

Read 2 more answers

if a carbon atom is bonded to two hydrogen atoms and two carbon atoms which type of bond must exist between the carbon atoms

zavuch27 [327]

Carbon to carbon bond is nonpolar covalent since the difference of their electronegativity’s is 0. Carbon has an electronegativity of 2.5. 2.5 - 2.5 is 0 making the bond between those two atoms nonpolar covalent. I hope I answered your question.

7 0

3 years ago

The temperature of 6.24 L of a gas is increased from 25.0°C to 55.0°C at constant pressure. The new volume of the gas is Questio

Sphinxa [80]

Answer:

Heating this gas to 55 °C will raise its volume to 6.87 liters.

Assumption: this gas is ideal.

Explanation:

By Charles's Law, under constant pressure the volume $V$ of an ideal gas is proportional to its absolute temperature $T$ (the one in degrees Kelvins.)

Alternatively, consider the ideal gas law:

$\displaystyle V = \frac{n \cdot R}{P}\cdot T$ .

$n$ is the number of moles of particles in this gas. $n$ should be constant as long as the container does not leak.
$R$ is the ideal gas constant.
$P$ is the pressure on the gas. The question states that the pressure on this gas is constant.

Therefore the volume of the gas is proportional to its absolute temperature.

Either way,

$V\propto T$ .

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}$ .

For the gas in this question:

Initial volume: $V_1 = \rm 6.24\; L$ .

Convert the two temperatures to degrees Kelvins:

Initial temperature: $T_1 = \rm 25.0\;\textdegree{C} = (25.0 + {\rm 273.15})\; K = 298.15\;K$ .
Final temperature: $T_1 = \rm 55.0\;\textdegree{C} = (55.0 + {\rm 273.15})\; K = 328.15\;K$ .

Apply Charles's Law:

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 6.24\;L \times \frac{328.15\; K}{298.15\;K} = 6.87\;L$ .

7 0

3 years ago