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2 years ago

3. Sand mixed with water

Chemistry

1 answer:

Alekssandra [29.7K]2 years ago

7 0

Answer:

<h3>Sand cannot mixed on water not float on water .</h3>

<h3>When we mix sand and water , No reaction take place . The sand simply settles down at the bottom of the water container . This is why sand is heavier than water and therefore cannot float in water .</h3>

<h2>Hope this helps you ✌️</h2>

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Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

7 0

3 years ago

Please Help! Let P and V represent the pressure and volume of Xe(g). If a piston is used to reduce the volume of the gas to V/2

Elza [17]

1) 2P

2) Average speed does not change

Explanation:

To solve the first part of the problem, we can use Boyle's Law, which states that:

"For a fixed mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume"

Mathematically:

$pV=const.$

where

p is the pressure of the gas

V is its volume

For the Xe gas in this problem we can write

$p_1 V_1 = p_2 V_2$

where:

$p_1 = P$ is the initial pressure

$V_1=V$ is the initial volume

$V_2=\frac{V}{2}$ is the final volume

Solving for p2, we find the final pressure:

$p_2=\frac{P V_1}{V_2}=\frac{pV}{V/2}=2P$

So, the pressure has doubled.

The average speed of the atoms/molecules in the gas is given by the formula

$v_{rms}=\sqrt{\frac{3RT}{M}}$

where

R is the gas constant

T is the absolute temperature (in Kelvin) of the gas

M is the molar mass of the gas

$v_{rms}$ is known as rms speed of the particles in the gas

From the formula, we see that the speed of the atoms in the gas depends only on the temperature of the gas.

In the Xe(g) gas in this problem, the temperature is kept constant; therefore, since nothing changes in the formula, this means that the average speed also does not change.

8 0

3 years ago

How much energy is released by the decay of 3 grams of 230Th in the following reaction 230 Th - 226Ra + "He (230 Th = 229.9837 g

Serhud [2]

<u>Answer:</u> The energy released for the decay of 3 grams of 230-Thorium is $2.728\times 10^{-15}J$

<u>Explanation:</u>

First we have to calculate the mass defect $(\Delta m)$ .

The equation for the alpha decay of thorium nucleus follows:

$_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}$

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

$\Delta m=(m_{Ra}+m_{He})-(m_{Th})$

$\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg$

(Conversion factor: $1amu=1.66\times 10^{-27}kg$ )

To calculate the energy released, we use Einstein equation, which is:

$E=\Delta mc^2$

$E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2$

$E=2.0916\times 10^{-13}J$

The energy released for 230 grams of decay of thorium is $2.0916\times 10^{-13}J$

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of = $2.0916\times 10^{-13}J$

Then, 3 grams of Th will release energy of = $\frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J$

Hence, the energy released for the decay of 3 grams of 230-Thorium is $2.728\times 10^{-15}J$

5 0

2 years ago

2. How many molecules are contained in 25 L of N₂ at S. T.P.?

irina [24]

Explanation:

How many nitrogen molecules are in 1 liter of nitrogen gas at STP?

Answer

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Pete Gannett

Ph.D. Chemistry, University of Wisconsin-Madison, (1982)2y

Seems to be an ideal gas law question. The relevant equation is:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles of gas, R is the gas constant (0.082 atm-L/mole-deg K), and T is temperature in Kelvins. STP means standard temperature and pressure and this is taken as 1 atm and 0º C or 273 K.

To calculate the number of molecules we will use the constant 6.023 * 10^23 molecules/mole and, therefore, we will need to know the number of moles (n). So, first we’ll rearrange the gas law equation, isolating ’n’ and then put the numbers in.

n = PV/RT = 1 * 1 / (0.082)(273) = 0.0447 moles

So, to calculate the number of molecules, multiple this by the number of molecules in a mole and you get:

# molecules of nitrogen in 1 Liter at STP = 6.023 * 10^23 molecules/mole * 0.0447 moles = 2.6905 * 10^22 molecules

Note, it does not matter what the gas is.

6 0

1 year ago

Read 2 more answers

How can you described the energy of a storm

lakkis [162]

Answer:

When the surface water is warm, the storm sucks up heat energy from the water, just like a straw sucks up a liquid. This creates moisture in the air. If wind conditions are right, the storm becomes a hurricane. This heat energy is the fuel for the storm.

Explanation:

I hope this can help you!

4 0

2 years ago

3. Sand mixed with water​

3. Sand mixed with water