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erastova [34]
1 year ago
11

Anna lives in a city that experiences high precipitation, with an average annual rainfall of 524 millimeters. It is warm all yea

r. It is hottest during July, with an average temperature of 98°F, and coldest during December, with an average temperature of 80°F. Tim lives in a city with an average rainfall of 30 millimeters per year. It has hot summers and cool winters. It is hottest during July, with an average temperature of 70°F, and coldest during January, with an average temperature of 28°F. Anna’s city is in a climate, while Tim’s city is in a climate.
Chemistry
1 answer:
gavmur [86]1 year ago
6 0

Anna lives in a city that is part of the tropical climate types. It has a constantly warm weather, and thus higher humidity, and according to the annual rainfall, it is most probably a rainfall that appears seasonally, not throughout the whole year.

Tim, on the other hand, lives in a city that is part of the dry climate types. It is most probably a place that is deep into the mainland, like the cold deserts of Central Asia, where the temperatures in the summer are high, and in winter are very low. Because of the distance from the sea, the rainfall doesn't reach this places, so they are very dry, and only have symbolic amount of annual rainfall.

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Find the mass in grams of 1.40x10^23 molecules of n2
mina [271]
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We calculate as follows:

</span>1.40x10^23 molecules of N2 ( 1 mol / 6.022 x 10^23 molecules ) ( 28.02 g / mol ) = 6.51 g N2
7 0
1 year ago
By mistake, a quart of oil was dumped into a swimming pool that measures 25.0 m by 30.0 m. The density of the oil was 0.750 g/cm
kondor19780726 [428]

The oil slick thick = 1.256 x 10⁻⁴ cm

<h3>Further explanation</h3>

Volume is a derivative quantity derived from the length of the principal

The unit of volume can be expressed in liters or milliliters or cubic meters

The conversion is

1 cc = 1 cm3

1 dm = 1 Liter

1 L = 1.06 quart

<em>so for 1 quart = 0.943 L</em>

\tt 0.943~L=0.943\times 10^{-3}m^3

Volume of oil dumped = volume of swimming pool

\tt 0.943\times 10^{-3}~m^3=25\times 30\times h(h=thick)\\\\h=\dfrac{0.943\times 10^{-3}}{750~m^2}=1.257\times 10^{-6}~m=\boxed{\bold{1.256\times 10^{-4}~cm}}

3 0
1 year ago
When an alpha particle is emitted from an unstable nucleus, the atomic mass numberof the nucleus
barxatty [35]

Answer:

The correct option is D.

Explanation:

Radioactive substances usually emit different types of particles when they are decaying. Such particles include alpha particles, beta particles and gamma ray. When an alpha particle is emitted from an unstable radioactive nucleus such nucleus usually lost an atomic mass that correspond to that of helium atom. Note that an alpha particle is made up of two protons and two neutrons, which result in mass number of 4. Thus, a nucleus that emit an alpha particle will have its mass number (atomic mass) reduce by 4 and atomic number that is reduced by 2.  

3 0
2 years ago
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
1 year ago
X-rays have a wavelength small enough to image individual atoms, but are challenging to detect because of their typical frequenc
Cerrena [4.2K]

Explanation:

Given: \lambda = 8.38\:\text{nm} = 8.38×10^{-9}\:\text{m}

Its frequency \nu is defined as

\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{8.38×10^{-9}\:\text{m}}

\:\:\:\:= 3.58×10^{16}\:\text{Hz}

7 0
1 year ago
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