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3 years ago

What is a dark mysterious and non-nicotine mineral containing sample of a very common silicate? It's geology actually.

Chemistry

1 answer:

Oxana [17]3 years ago

4 0

Orthoclase & Plagioclase...

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6. 5.50 x 10- molecules of carbon dioxide to moles.

sergeinik [125]

1.24973017189471 is probably the answer to your equation

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3 years ago

For the reaction, 2SO2(g) + O2(g) <--> 2SO3(g), at 450.0 K the equilibrium constant, Kc, has a value of 4.62. A system wa

Nady [450]

Answer:

To the left.

Explanation:

Step 1: Write the balanced reaction at equilibrium

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

Step 2: Calculate the reaction quotient (Qc)

Qc = [SO₃]² / [SO₂]² × [O₂]

Qc = 0.254² / 0.500² × 0.00855

Qc = 30.2

Step 3: Determine in which direction will proceed the system

Since Qc > Kc, the system will shift to the left to attain the equilibrium.

4 0

3 years ago

Can you pls answer this question cuz i don't know what is the answer on this..

Marysya12 [62]

Answer:

here's is ur answer in attachment

Explanation:

mark it brainliest if it helps you ❤️

3 0

3 years ago

How many atoms are in NH4Cl.

RSB [31]

Answer:

there are 6

Explanation:

3 0

3 years ago

The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p

saul85 [17]

Answer : The vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of bromine at $10.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $10.0^oC=273+10.0=283.0K$

$T_2$ = normal boiling point of bromine = $58.8^oC=273+58.8=331.8K$

$\Delta H_{vap}$ = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})$

$P_1=0.1448atm$

Hence, the vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

4 0

3 years ago