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sweet-ann [11.9K]

4 years ago

5

the coast of California on the western side of the state has a voice climate wild Eastern side of the desert.what could cause th

is diffrence

2 answers:

diamong [38]4 years ago

7 0

fires could cause a big difference in the weather of California

Elanso [62]4 years ago

7 0

I think the answer is B, Mountains prevent moist air from blowing over into the desert.

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What isotope is the product of the alpha (*2He) nuclear decay of Radon-222 (222U)?

NikAS [45]

Answer:

ⁿₐX => ²¹⁸₈₄Po

Explanation:

Let ⁿₐX be the isotope.

Thus, the equation can be written as follow:

²²²₈₆Rn —> ⁴₂α + ⁿₐX

Next, we shall determine the value of 'n' and 'a'. This can be obtained as follow:

222 = 4 + n

Collect like terms

222 – 4 = n

218 = n

Thus,

n = 218

86 = 2 + a

Collect like terms

86 – 2 = a

84 = a

Thus,

a = 84

ⁿₐX => ²¹⁸₈₄Po

²²²₈₆Rn —> ⁴₂α + ⁿₐX

²²²₈₆Rn —> ⁴₂α + ²¹⁸₈₄Po

3 0

3 years ago

If a substance has a mass of 12.50 g and takes up 3.4 mL of space, what is the density?

miskamm [114]

Answer:

The answer is

<h2>3.68 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3> $density = \frac{mass}{volume}$ </h3>

From the question

mass of substance = 12.50 g

volume = 3.4 mL

The density of the substance is

$density = \frac{12.50}{3.4} \\ = 3.676470588...$

We have the final answer as

<h3>3.68 g/mL</h3>

Hope this helps you

7 0

3 years ago

Срочноооооооооооооо!!!!

Ymorist [56]

Answer:

I can't answer this as I am not able to read the Question

4 0

3 years ago

Write the congruence condition in symbolic form for each pair of triangles in pic above. If wrong answer I will report. If corre

arsen [322]

Answer: ABC&AMN are congruent by ASA comgruence

angle A=A

sideAN=AB

angle M=C

Explanation:

4 0

3 years ago

Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)

Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation Cr(s) → Cr3+ + 3e-

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the half-reactions

2(Cr → Cr3+ + 3e-) E° ox = 0.74 V

3(Ni2+ +2e- → Ni) E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox = -0.25 + 0.74 = 0.49

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

b)

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution) MnO4- +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e- (oxidation) Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the half-reactions

MnO4- +8H+ +10-e- ⇔ Mn2+ +4H2O E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e- E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox = 1.51 + 2.12 = 3.63 V

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0

3 years ago