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2 years ago

Please help will mark brainliest!

Chemistry

1 answer:

Llana [10]2 years ago

8 0

Answer:

1. desert

2. grasslands

3. rainforest

4. tundra

5. temperate deciduous forest

6. coniferous forest

Explanation:

educated guess, good luck

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At what temperature and pressure can all three phases of water exist simultaneously?

Rzqust [24]

Thirty seven I think

8 0

3 years ago

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Balance this equation: Al+ HNO3 ---- H2 + Al(NO3)3

romanna [79]

Answer:

2Al+ 6HNO3 ---- 3H2 + 2Al(NO3)3

Explanation:

Put coefficient a,b,c, and d for calculation:

a Al + b HNO3 = c H2 + d Al(NO3)3

for Al: a = d

for H: b = 2c

for N: b = 3d

for O: 3b = 9d

Suppose a=1, then d=1, b=3, c=3/2

multiply 2 to make all natural number, a=2, then b=6, c=3, d=2

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3 years ago

Suppose that 4.8 L of methane at a pressure

Ghella [55]

Answer:

972.3 Torr

Explanation:

P2=P1V1/V2

You can check this by knowing that P and V at constant T have an an inverse relationship. Hence, this is correct.

6 0

2 years ago

What does lithium 6 and lithium 7 look like

Marysya12 [62]

Lithium 6 would have 6 valence electrons in the outer orbital, while lithium 7 would have 7 in the outer orbital.

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2 years ago

When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at

STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

$1.22 * 10^{5} = - 8.314* 2400 * ln K$

$\\ 1.22 * 10^{5} = -19953.6 * ln K$

$ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022$

So, the equilibrium constant is 0.0022.

4 0

2 years ago