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2 years ago

How many moles of phosphorus trichloride are in 200 grams of the molecule???

Physics

1 answer:

sesenic [268]2 years ago

8 0

<h3>Answer:</h3>

1.46 moles

<h3>Explanation:</h3>

<u>We are given</u> ;

Mass of phosphorus trichloride is 200 g

We are required to calculate the number of moles of phosphorus trichloride;

We know that to calculate the number of moles we divide mass by the molar mass.

That is;

Moles = Mass ÷ Molar mass

Molar mass of phosphorus trichloride = 137.33 g/mol

Therefore;

Moles = 200 g ÷ 137.33 g/mol

= 1.456 moles

= 1.46 moles

Therefore, the moles of phosphorus trichloride are 1.46 moles

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What did Thomson’s and Rutherford’s experiments have in common? They both used charged particles in their experiments. They both

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Answer:

Both Thomson and Rutherford used charged particles in their experiments.

Explanation:

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2 years ago

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the moderate temperatures of islands throughout the world has much to do with water's vast supply of internal energy. high evapo

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High specific heat of the water. Option (c)

What is Specific heat?

The amount of heat required to increase the temperature of one gram of a substance by one degree Celsius is referred to as the substance's specific heat. Typically, calories or joules are used per gram and degree Celsius when referring to the units of specific heat.

The moderate temperature of islands has much to do with the water's high specific heat. The typical off-water is more significant than this clear land or soil. Due to this fact, water absorbs and releases eat more slowly. In comparison to the land.

Hence, the water has high specific heat.

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7 months ago

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

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2 years ago

Please Please Please Can Help Me On This Question!!!!! I Give Thanks!!!! Please Do 1-4!!!!

ololo11 [35]

1.add the amount of the diagram which is M+Y then dived the answer you get.

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2 years ago

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As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

$-F_3sin(b) + F_1 = 0$

For the address and we have:

$-F_3cos(b) + F_2 = 0$

The forces $F_1$ and $F_2$ are known

$F_1 = 5.7\ N\\\\F_2 = 1.9\ N$

We have 2 unknowns ( $F_3$ and b) and we have 2 equations.

Now we clear $F_3$ from the second equation and introduce it into the first equation.

$F_3 = \frac{F_2}{cos (b)}$

Then

$-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°$

Then we find the value of $F_3$

$F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N$

Finally the answer is 6.3 N at 162° counterclockwise from

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2 years ago