Ask question

Ask question

All categories

3 years ago

6

How are Celsius, Kelvin, and Fahrenheit scales similar? How are they different? Considering their differences, think of one scen

ario in which each scale is more convenient to use?

1 answer:

PIT_PIT [208]3 years ago

6 0

Answer:

they are similar becauze they measure temperature

You might be interested in

A tank holds 6.90 m3 of water. What is the volume in ft3? (3.28 ft = 1 m, Remember significant figures, but this answer does not

AleksAgata [21]

Volume of tank = $6.9 m^{3}$ (given)

Since, $1 m = 3.28 ft$

So,

$1 m^{3} = 35.287 ft^{3}$

For $6.9 m^{3}$ :

$6.9\times 35.287 ft^{3} = 243.4803 ft^{3}$

The significant rule for multiplication, states that the number of significant figures in the answer obtained by multiplication is determined by the value with the lowest number of significant digits.

Since, the minimum number of decimal places in the above multiplication operation is 1 so, the final result must be upto 1 decimal place only.

$243.4803 ft^{3} \simeq 243.5 ft^{3}$

Hence, volume in $ft^{3}$ is 243.5.

5 1

2 years ago

Read 2 more answers

A pure substance that is made of only one kind of atom is called _________. A. an element B. an electron C. a particle D. a prot

tekilochka [14]

An element is a pure substance made up of only one kind of atom. An element is composed of atoms in which it is composed of neutrons and protons. The nucleus is surrounded by electrons. The answer to this problem is A. element

5 0

3 years ago

Read 2 more answers

In a unimolecular reaction with twice as much starting material as product at equilibrium, what is the value of Keq? Is ΔG o pos

lubasha [3.4K]

Answer : The value of $K_{eq}$ is, 0.5 and $\Delta G$ = positive.

Explanation :

The unimolecular reaction is:

$A\rightarrow B$

In unimolecular reaction, the starting material is 2 times to the product.

$A=2B$ .........(1)

As we know that:

$K_{eq}=\frac{B}{A}$ ...........(2)

Now substitute equation 1 in 2, we get:

$K_{eq}=\frac{\frac{A}{2}}{A}$

$K_{eq}=0.5$

Now we have to calculate the value of $\Delta G^o$ at 298 K.

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = gas constant = 8.314 J/mol.K

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = 0.5

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

$\Delta G^o=-(8.314J/mol.K)\times (298K)\ln (0.5)$

$\Delta G^o=1717.32J/mol$

Thus, the value of $\Delta G^o$ at 298 K is, 1717.32 J/mol

As we know that:

$\Delta G$ = +ve, reaction is non spontaneous

$\Delta G$ = -ve, reaction is spontaneous

$\Delta G$ = 0, reaction is in equilibrium

Thus, the $\Delta G$ = +ve. So, the reaction is non spontaneous.

8 0

3 years ago

Find the molar mass of AI(NO3)3?<br><br> Explain plz!

NeX [460]

Answer:

212.996 or 213 approx.

hope it helps.

Molar mass of Al - 27

Molar mass of N - 14

Molar mass of O - 16

so, the molar mass of (No3)3 will be (14+16×3)3 which is equals to 186.

so, Molar mass of Al(No3)3 will be 186+27 = 213.

8 0

2 years ago

Read 2 more answers

Weather balloon is filled with 62.3 L of helium gas at 760 mm of Hg after being released the volume increases to 130 L as the ba

enot [183]

Answer : The pressure at this new altitude is, 364.2 mmHg

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 760 mmHg

$P_2$ = final pressure = ?

$V_1$ = initial volume = 62.3 L

$V_2$ = final volume = 130 L

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

$760mmHg\times 62.3L=P_2\times 130L$

$P_2=364.2mmHg$

Thus, the pressure at this new altitude is, 364.2 mmHg

7 0

3 years ago