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3 years ago

When mud and water are combined, the result is ______.

Chemistry

2 answers:

Nimfa-mama [501]3 years ago

7 0

A. A mixture

You are mixing both mud and water so a mixture

Arturiano [62]3 years ago

4 0

Answer:

A. a mixture

Explanation:

hope this helps

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How many moles of neon atoms are there in a neon sign that has 2.4 * 20^ 24 atoms of neon? *

Whitepunk [10]

Answer:

4 moles of neon

Explanation:

Given data:

Number of moles of neon = ?

Number of atoms of neon = 2.4×10²⁴ atoms

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For given neon atoms:

1 mol = 6.022 × 10²³ atoms

2.4×10²⁴ atoms × 1 mol / 6.022 × 10²³ atoms

0.4×10¹ mol = 4 mol

3 0

3 years ago

a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a bu

Alex17521 [72]

Answer:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

3 0

3 years ago

How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of

alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

$E = k \frac{q}{r^{2} }$

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

$E = k \frac{q}{r^{2} }$

q = E x r²

q = 1250 N/C x 0.205m²

8.99 x 10⁹ Nm²/C²

q = 5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

$n = \frac{q}{e}$

n = 5.84 x 10⁻⁹ C

1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

3 0

3 years ago

Do you think lava or water formed the Chanel on Mars? And why? I’ll give brainliest

Kamila [148]

Answer:

A channel in an ancient Martian "river bed" was not carved out by liquid water but built by molten lava

6 0

2 years ago

If the E°cell value = +0.46V and E°(Ag+/Ag) = +0.80V

fenix001 [56]

Answer:

0.34

Explanation:

Cu>>>Cu2+(oxidation number increases)[Oxidation occurs]

Ag+>>>Ag(Oxidation number decreases) [Reduction occurs]

E°=0.80-0.46

=0.34V

8 0

2 years ago

Read 2 more answers