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Cloud [144]
3 years ago
6

Seawater is composed of salt, sand, and water. is seawater a heterogeneous or homogeneous mixture? explain

Chemistry
1 answer:
Umnica [9.8K]3 years ago
4 0

A mixture is said to be homogeneous if its composition is uniform throughout the mixture. They are referred to as solutions. On the other hand, heterogeneous mixture does not have uniform composition. The substances present in the mixture have visible difference or phases.

Sea water is composed of salt, sand and water. Here, salt and water form homogeneous mixture but due to the presence of sand, the mixture is heterogeneous. Salt dissolves in water and the solution has uniform composition but sand does not dissolve in water, even after vigorous mixing, after some time it settles at the bottom resulting formation of layers. Thus, sea water containing salt, water and sand is a heterogeneous mixture.


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a cylinder rod formed from silicon is 46.0 cm long and has a mass of 3.00 kg. the density of silicon is 2.33 g/cm³. what is the
Gnesinka [82]
d=\frac{m}{V}\\\\
V=\pi r^{2}H\\d=\frac{m}{\pi r^{2}H} \ \ \ |*\pi r^{2}H\\\\
d \pi r^{2}H=m \ \ \ |:d \pi H\\\\
r^{2}=\frac{m}{d \pi H}\\\\
r=\sqrt{\frac{m}{d \pi H}}\\\\
2r=d\\\\
d=2\sqrt{\frac{m}{d \pi H}}

H=43cm\\
d=2.33\frac{g}{cm^{3}}\\
m=3kg=3000g\\\\
d=2\sqrt{\frac{m}{d \pi H}}=2*\sqrt{\frac{3000g}{2.33\frac{g}{cm^{3}} * \pi * 43cm}}=2*\sqrt{\frac{3000}{100,19\pi \frac{1}{cm^{2}}}}\approx6,17cm
7 0
3 years ago
Diethyl ether is produced from ethanol according to the following equation:
juin [17]
The balanced chemical equation is given as:

2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)

We are given the yield of  CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.

Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3 

% yield = .775 = actual yield / 264.66


actual yield = 205.11 g CH3CH2OCH2CH3
6 0
2 years ago
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8 0
3 years ago
What is the new concentration of 25.0mL added to 125.0mL of 0.150M
____ [38]

Answer:

The new concentration is 0.125 M.

Explanation:

Given data:

Initial volume V₁ = 125.0 mL

Initial molarity M₁ = 0.150 M

New volume V₂ = 25 mL +125 mL = 150 mL

New concentration M₂ = ?

Solution:

M₁V₁    =    M₂V₂

0.150 M × 125 mL = M₂ × 150 mL

M₂ = 0.150 M × 125 mL / 150mL

M₂ = 18.75 M.mL/150 mL

M₂ = 0.125 M

The new concentration is 0.125 M.

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