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Paul [167]
3 years ago
6

How many Br atoms are in 1.98 g of Br?

Chemistry
1 answer:
Mkey [24]3 years ago
7 0

Answer:

1.5 x 10²²atoms

Explanation:

Given parameters:

Mass of Br = 1.98g

Unknown:

Number of atoms

Solution:

A mole of a substance is the amount of substance contained in the avogadro's number of particles.

To solve this problem, we must first find the number of moles present in the given mass of the Br atom:

       Number of moles = \frac{mass}{molar mass} =  \frac{1.98}{80}  = 0.025mole.

Now we know that:

       1 mole = 6.02 x 10²³

      0.025 mole = 0.025 x 6.02 x 10²³ = 1.5 x 10²²atoms

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How many grams are there in 3.3 x 1023 particles of N2I6?
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Answer: With the exception of hydrogen, all atoms have three main parts. The parts of an atom are protons, electrons, and neutrons. A proton is positively charged and is located in the center or nucleus of the atom. Electrons are negatively charged and are located in rings or orbits spinning around the nucleus.

Explanation:

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3 years ago
A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release
Marrrta [24]

Answer:

V_2=12.1L

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:

T_1=20+273=293K\\\\T_2=-40+273=233K

Then, we obtain:

V_2=\frac{P_1V_1T_2}{T_1P_2}\\\\V_2=\frac{0.987atm*8.50L*233K}{293K*0.550atm}\\\\V_2=12.1L

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5 0
3 years ago
If the mass of the object below is 28g, what is the density of the object below. Units are in cm below. Please round your answer
Murljashka [212]

Answer:

d = 0.93 g/cm³

Explanation:

Given data:

Mass of object = 28 g

Volume of object = 3cm×2cm×5cm

density of object = ?

Solution:

Volume of object = 3cm × 2cm ×5cm

Volume of object = 30 cm³

Density of object:

d = m/v

by putting values,

d = 28 g/ 30 cm³

d = 0.93 g/cm³

3 0
3 years ago
Give all possible ml values for orbitals that have each of the following: (a) l = 2; (b) n = 1; (c) n = 4, l = 3.
Olin [163]

Answer:

(a) ml = 0, ±1, ±2

(b) ml = 0

(c) ml = 0, ±1, ±2, ±3, ±4

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Subshell number, 0 ≤ l ≤ n − 1

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So in our exercise,

(a) l = 2; equivalent with with sublevel <em>d</em>

-l ≤ ml ≤ l, ml = 0, ±1, ±2, equivalent with dxy, dxz, dyz, dx2-y2, dz2

(b) n = 1;

n = 1, only 01 level

l = 0, equivalent with sublevel <em>s</em>

ml = 0

(c) n = 4, l = 3.

l = 3, equivalent with sublevel <em>f</em>

ml = 0, ±1, ±2, ±3, ±4

7 0
3 years ago
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