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lora16 [44]
2 years ago
8

Try the same filtering method with the salt water. What happened? Taste the

Chemistry
2 answers:
enyata [817]2 years ago
3 0

Yeah, what? I don't think Brainly knows that. Maybe you should taste the filtered salt water and find out the answer.

Lerok [7]2 years ago
3 0

Answer:

It should, well, taste less salty or even like fresh water if it is filtered enough

Explanation:

Depending on how you filter it, the salt should separate out of the water

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Where are elements with similar chemical properties found on the periodic table?
LekaFEV [45]
All the members of a family of elements have the same number of valence electrons and similar chemical properties. The horizontal rows on the periodic table are called periods.
8 0
3 years ago
A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h
frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]

Where;

\frac{f_2}{f_1}  is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1}  = 10^{0.7289}\\\\\frac{f_2}{f_1}  = 5.356

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0
3 years ago
Choose the description that matches the law of included fragments.
GREYUIT [131]

Answer:

The answer is B

Explanation:

3 0
2 years ago
A 45.0 g sample of a metal at 85.6 °C is placed in 150.0 g of water at 24.6 °C. The final temperature of the system is 28.3 °
earnstyle [38]

Answer:

904.014 j/kgk

Explanation:

Mass of metal = 45g

Temperature of metal = 85.6°

Mass of water = 150

Temperature of water = 24.6

Final temperature of system = 28.3

Heat lost by metal = Heat gained by water

m1 * c1 * dt = m2 * c2 * dt

Q = quantity of heat

Q = m*c*dt

dt = change in temperature

dt of water = 28.3 - 24.6 = 3.7

dt of metal = 85.6 - 28.3 = 57.3

Specific heat capacity of water, c = 4200

(45 * 10^-3) * c * 57.3 = (150 * 10^-3) * 4200 * 3.7

2.5785c1 = 2331

c1 = 2331 / 2.5785

= 904.01396

= 904.014 j/kgk

3 0
2 years ago
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Answer:

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3 0
3 years ago
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