All categories

3 years ago

Which of the following is not a valid equation for describing the behavior of

Chemistry

2 answers:

bonufazy [111]3 years ago

5 0

D is the answer!!!!!!!!!!!!!!!!

AnnZ [28]3 years ago

3 0

Answer:

A. (P1/V1) = (P2/V2) is not a valid equation

D. (V2/T1) =V2/T2) is not a valid equation. ( V1/T1) = (V2/T2) is a valid equation

Explanation:

Step 1: Data given

The gas law is p*V = nRT (this means option B is a valid equation)

If we compare 2 gases we can write this as:

p1*V1 / ( n1*R*T1 ) = p2*V2 / ( n2*R*T2 )

Step 2: When temperature and number of moles are constant

This means volume and pressure will change.

p1*V1 / ( n1*R*T1 ) = p2*V2 / ( n2*R*T2 ) OR p1¨V1 = p2*V2

The relationship for Boyle’s Law can be expressed as follows: p1V1 = p2V2, where p1 and V1 are the initial pressure and volume values, and p2 and V2 are the changed values of the pressure and volume of the gas.

When the initial pressure rises, the initial volume will decrease (and vice versa). For a fixed mass of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional.

For the formula (P1/V1) = (P2/V2) it would mean pressure and volume are directly proportional, this is not true. So option A is not a valid equation.

Step 3: When pressure and number of moles are constant

This means volume and temperature will change.

p1*V1 / ( n1*R*T1 ) = p2*V2 / ( n2*R*T2 ) OR V1/T1 = V2/T2

For the formula (V1/T1) =V2/T2) it would mean volume and temperature are directly proportional. This is a valid equation

Option D would only be a valid equation if its (V1/T1) =V2/T2)

(V2/T1) =V2/T2) is not a valid equation. ( I don't know if this was a thypo or not).

You might be interested in

The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac

creativ13 [48]

Answer:

a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$ c) No $H_{2}$ was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$

c) $H_{2}$ was not the limiting reactant based on the mol to mol ratio of $H_{2}$ and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of $H_{2}$ would also be produced.

4 0

3 years ago

How many moles of Cu(OH)2 will be produced if given 1 mole KOH.

stiks02 [169]

Answer:

0.032 mole

Explanation:

Please mark me as brainliest

7 0

1 year ago

A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu

bogdanovich [222]

Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

Explanation:

The temperature of an object can be written using different temperature scales.

The two most important scales are:

- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).

- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

$T(K)=T(^{\circ}C)+273.15$