All categories

3 years ago

Why are there sometimes warm days during a cold winter? (not chemistry, its actually science)

Chemistry

1 answer:

nika2105 [10]3 years ago

4 0

Answer:

It is all about the tilt of the Earth's axis. Many people believe that the temperature changes because the Earth is closer to the sun in summer and farther from the sun in winter. In fact, the Earth is farthest from the sun in July and is closest to the sun in January!

You might be interested in

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

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

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

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

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

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

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago

Which prefix is used in the name of a stereoisomer to indicate that the

tester [92]

Answer:

<em>"cis" (Latin for Side) </em>

3 0

2 years ago

What is the ratio of carbon, hydrogen, and oxygen in most carbohydrates?

horrorfan [7]

In carbohydrates the C:H:O is 1:2:1

7 0

3 years ago

Read 2 more answers

10. Hydrogen peroxide (H2O2: M = 34 g mol-1) is a powerful oxidising agent that is used in concentrated solution in rocket fuel

Lubov Fominskaja [6]

Answer:

The molality of solution=12.605 m

Explanation:

We are given that

Molar mass of Hydrogen peroxide, M=34 g/mol

Density of solution, $\rho=1.11gcm^{-3}$

30% Means mass of solute (Hydrogen peroxide)=30 g

Mass of solvent =100-30=70 g

Total mass of solution, m=100 g

Number of moles of solute= $\frac{given\;mass}{molar\;mass}$

Using the formula

Number of moles of hydrogen peroxide= $\frac{30}{34}$

Now, molality of solution

$m=\frac{number\;of\;moles\;of\;solute}{mass\;of\;solvent}\times 1000$

$m=\frac{30}{70\times 34}\times 1000$

$m=12.605 m$

Hence, the molality of solution=12.605 m

6 0

3 years ago

How many moles are in 4.35 X 1024 atoms of Iron (Fe)?

Naya [18.7K]

Answer:

<h3>The answer is 7.23 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{4.35 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 7.225913...$

We have the final answer as

<h3>7.23 moles</h3>

Hope this helps you

3 0

3 years ago