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avanturin [10]
3 years ago
13

Evaluate the following conversion. Will the answer be correct? Explain. If incorrect, how could you adjust one of the factors to

arrive at a rate?
rate=75 m1 s×60 s1 min×1 h60 m
Chemistry
1 answer:
elena55 [62]3 years ago
6 0

Given :

Rate = 75 m 1 s×60 s 1 min×1 h 60 m.

To Find :

Correct answer after conversion.

Solution :

We know, 1 min = 60 sec.

1 hour = 60 min = 60×60 sec = 3600 sec.

Putting value of min and hour in seconds , we get :

R=((75\times 60) + 1 )\times ( (60 + 1)\times 60 ) \times ( 3600+3600)\ s^3\\\\R=1.186\times 10^{11} \ s^3

Hence, this is the required solution.

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An ideal gas sample is confined to 3.0 L and kept at 27 °C. If the temperature is raised to 77 °C and the initial pressure was 1
Nadusha1986 [10]

The gas is confined in 3.0 L container ( rigid container) ⇒ the volume remains constant when the temperature is increased from from 27oC to 77oC and therefore V1=V2 .

<h2>Hope it helps you please mark as brainlist</h2>

6 0
3 years ago
Calculate the mass of calcium chloride that contains 3.20 x 1024 atoms of chlorine.
frez [133]

Answer:

294.87 gm  CaCl_2

Explanation:

The computation of the  mass of calcium chloride is shown below:

But before that following calculations need to be done

Number of moles of chlorine atom is

= 3.20 × 10^24 ÷ 6.022 × 10^23

= 5.314 moles

As we know that

1 mole CaCl_2 have the 2 moles of chlorine atoms

Now 5.341 mole chloride atoms would be

= 1 ÷ 2 × 5.314

= 2.657 moles

Now

Mass of CaCl_2 = Number of moles × molar mass of  CaCl_2

= 2.657 moles × 110.98 g/mol

= 294.87 gm  CaCl_2

7 0
3 years ago
When particles are removed volume will do what
DIA [1.3K]
Volume will decrease over time, the more particles removed, is making the volume decrease. 
7 0
3 years ago
The _________ __________ is the bottom number on each element?
kati45 [8]
The full chemical symbol for an element<span> shows its mass </span>number<span> at the top, and its atomic </span><span>number at the bottom</span>

7 0
3 years ago
g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig
kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, \Delta S_{env} = -1.18 J/K

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

6 0
2 years ago
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