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2 years ago

9

State your claim. Make sure your claim fully explains how you can distinguish between materials that appear similar. HELPPPPPP M

EEEEE PLZZZZZZZZ!!!!!!!!!!!!!!!!!!!!!

2 answers:

snow_lady [41]2 years ago

7 0

The right one is darker than the left

asambeis [7]2 years ago

5 0

The material on the right is a darker shade than the one on the left. Also the one on the right also is less of a block and a little more shagged

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The volume of a gas at 17.5 psi decreases from 1.8L to 750mL. What is the new pressure of the gas in arm?

Maurinko [17]

Answer:

P₂ = 2.88 atm

Explanation:

Given data:

Initial volume of gas = 1.8 L

Final volume = 750 mL

Initial pressure = 17.5 Psi

Final pressure = ?

Solution:

We will convert the units first:

Initial pressure = 17.5 /14.696 = 1.2 atm

Final volume = 750 mL ×1L/1000L = 0.75 L

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

1.2 atm × 1.8 L = P₂ ×0.75 L

P₂ = 2.16 atm. L/ 0.75 L

P₂ = 2.88 atm

4 0

3 years ago

Why do experiments need to be repeatable?

maw [93]

If they were not repeatable people would think the experiment is not accurate. If it can be repeated than the data can prove a very valid point.

3 0

3 years ago

Read 2 more answers

Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. T

Ostrovityanka [42]

Explanation:

The given data is as follows.

$\Delta H$ = 286 kJ = $286 kJ \times \frac{1000 J}{1 kJ}$

= 286000 J

$S_{H_{2}O} = 70 J/^{o}K$ , $S_{H_{2}} = 131 J/^{o}K$

$S_{O_{2}} = 205 J/^{o}K$

Hence, formula to calculate entropy change of the reaction is as follows.

$\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)$

= $[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]$

= $[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]$

= 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

$\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}$

= $286000 J - (163.5 J/K \times 298 K)$

= 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

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4 years ago

What is a homogeneous ?

USPshnik [31]

Consisting of parts all of the same kind material.

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3 years ago

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A sample of gas occupies 3.00 L with 5.00 moles present. What would

pantera1 [17]

3/5 times 5/3x = 8*3/5. X=24/5 simplified would be x= 4.8 L.

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3 years ago