PLS I M IN HURRY.

Serhud [2]

Let the mass be X g

percentage = X/ 6.50 * 100 =2.2%

X= 0.143 g

The mass is 0.143 g

8 0

3 years ago

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..can someone answer these questions ive already posted this question 16 times and I dont have much points left so please anyone

noname [10]

Answer

A
D
D
A
B
A

This is what I got, but i'm mot sure if I'm right

7 0

3 years ago

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During photosynthesis the chemical bonds between are broken

Vaselesa [24]

Answer:

yes they are broken down into smaller units

5 0

2 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

A 150.0mL sample of an aqueous solution at 25C contains 15.2mg of an unknown nonelectrolyte compound. If the solution has an osm

Nookie1986 [14]

Answer:

MM = 225.11 g/mol

Explanation:

In this case, let's analyze the data.

We have a 15.2 mg of an unknown electrolyte in 150 mL of solution. The osmotic pressure is 8.44 Torr, and we want the molecular mass of the unknown.

The osmotic pressure can be calculated using the following expression:

π = CRT (1)

π: osmotic pressure (8.44 Torr * 1 atm / 760 Torr = 0.011 atm)

C: Concentration of the unknown in the solution

R: universal constant of gases (0.082 L atm / K mol)

T: temperature in Kelvin (25 + 273 = 298 K)

From this expression, we can either solve for C, and then use another expression to calculate the molecular mass, or we can just replace the expressions in the above formula, to get the direct molecular mass. In this case, we'll follow the second method.

Concentration or molarity of a substance can be calculated using:

C = moles / V (2)

And moles can be calculated using this expression:

moles = m / MM (3)

Replacing (3) in (2), and then in (1) we have:

C = m / MM * V

π = m * RT / MM * V (4)

and now, we can solve for MM:

MM = mRT / π V (5)

Now, we just need to replace the given data into the above expression to get the value of the molecular mass:

MM = (15.2 / 1000) * 0.082 * 298 / 0.011 * 0.150

Hope this helps

3 0

3 years ago