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Westkost [7]
2 years ago
5

A student accidentally left a bottle of ammonia unclosed in the corner of a laboratory. After a few minutes the entire laborator

y smelled of ammonia. Which process has occurred?
Diffusion
Distillation
Evaporation
Brownian motion
Chemistry
2 answers:
Artyom0805 [142]2 years ago
4 0
Diffusion I’d the spreading of something more widely.

The smell of ammonia diffused throughout the laboratory.

Answer : Diffusion
alexdok [17]2 years ago
3 0
The locket necklace symbolizes Love. Evident from how she was feeling as he held her hand.
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The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]

In NH3 , let oxidation number of N be x

x + (+1)3 = 0

x = -3

In HNO3 , let oxidation number of N be x

1 + x + (-2)3 = 0

x = +5

In NO2 , let oxidation number of N be x

x + (-2)2 = 0

x = +4
5 0
2 years ago
A chemist pours 1 mol of zinc granules into one beaker and 1 mol of zinc chloride powder into another beaker. What do the two sa
Oksana_A [137]
Answer is: <span> two samples have in common same amount of substance and same number of particles.
1) There are same amount of substance in both beakers:
n(Zn) = 1 mol.
n(ZnCl</span>₂) = 1 mol.
2) There are same number of particles (atoms, molecules, ions) in both beakers:
N(Zn) = n(Zn) · Na.
N(Zn) = 1 mol · 6.023·10²³ 1/mol = 6.023·10²³ atoms of zinc.
N(ZnCl₂) = n(ZnCl₂) · Na.
N(ZnCl₂) = 1 mol · 6.023·10²³ 1/mol = 6.023·10²³ molecules of zinc(II) chloride.
Na - Avogadro number.


5 0
2 years ago
Read 2 more answers
Suggest why sodium and hydrogen ions do not diffuse at the same rate
Troyanec [42]

Answer:

sodium has got ionic bonds that are weak

compared to hydrogen covalent bonds that are strong

8 0
2 years ago
In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?
vladimir2022 [97]

Answer:

The volume of the container is 59.112 L

Explanation:

Given that,

Number of moles of Oxygen, n = 3

Temperature of the gas, T = 300 K

Pressure of the gas, P = 1.25 atm

We need to find the volume of the container. For a gas, we know that,

PV = nRT

V is volume

R is gas constant, R =  0.0821 atm-L/mol-K

So,

V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L

So, the volume of the container is 59.112 L

6 0
2 years ago
Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter
Margarita [4]

Answer:

thickness = 0.29 cm

Explanation:

In order to make fake iron ball [made of gold] we should get mass of fake ball

should be equal to that of Iron ball.So for that we should calculate

volume of iron ball using diameter given ;formula is 4/3 pi r^3

given d= 6 cm;so radius r= 6/2 = 3 cm

then volume of Iron ball = 4/3 *3.14* 3^3 = 113.04 cm^3

So mass of iron ball = volume x density = 113.04 * 5.15 g/cm^3 = 582.156 g

This must be the mass of gold ball ;now calculate volume of gold ball using its

density

volume of gold ball = mass of gold ball/density of gold ball = 582.156 g/19.3 g/cm^3

= 30.1635 cm^3

Now this must be the volume of hollow sphere whose outer radius R = 3cm

and inner radius r= ??

Volume of hollow ball = 4/3pi[R3-r^3]

30.1635 cm^3 = 4/3 pi [3^3-r^3]

30.1635* 3/4*3.14 = 27 - r^3

7.2046 = 27- r^3

r^3 = 19.7954

r= 2.7051 cm

So the thick ness = outer radius- inner radius = 3 - 2.7051 = 0.2949 cm

rounding to 2 significant figures

we get thickness = 0.29 cm

8 0
2 years ago
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