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lesantik [10]
4 years ago
11

The formation of glaciers is directly controlled by climate. True False

Chemistry
2 answers:
skad [1K]4 years ago
8 0
True! Hope this helps:)
Agata [3.3K]4 years ago
5 0
The answer I am pretty sure is true
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Compared to most substances, water is unusual because it ____ when it goes from the liquid to solid state.
yuradex [85]

Answer:

expands

Explanation:

Most substances become increasingly compressed when frozen solid, but water is a famous exception.

6 0
3 years ago
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Can someone help me with this please!
pashok25 [27]
Nit<u>ITE</u><u /> is NO<u>2</u><u />
nitr<u>ATE</u><u /> is NO<u>3</u><u />
5 0
3 years ago
Problem PageQuestion Gaseous ethane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of w
Nata [24]

Answer:

<h2> = ( 1.08 / 2.2 ) 100% = 49%</h2>

Explanation:

Balanced Equation: 2CH₃CH₃(g) + 5O₂(g) → 2CO₂(g) + 6H₂O(g)

Calculate moles of CH₃CH₃ and O₂

1.2 ₃₃ ( 1 ₃₃/ 30.0694 ₃₃ ) = 0.040 ₃₃

8.6 ₂ ( 1 2/ 31.998 ₂ ) = 0.27 ₃₃

Find limiting reagent  0.040 ₃₃ ( 5 ₂/ 2 ₃₃ ) = 0.10 ₂

CH₃CH₃ is the limiting Reagent

CH₃CH₃ (L.R.) O₂ CO₂ H₂O

Initial (mol) 0.040 0.27 0 0

Change (mol) -2x=-0 -5x= -0.10  +2x=+0.040 +6x=+0.12

Final (mol) 0 0.117 0.040 0.12

0.040 − 2 = 0 = 0.020

Determine percent yield

0.12 ₂ ( 18.0148 ₂ /1 ₂ ) = 2.2 ₂  

= ( 1.08 / 2.2 ) 100% = 49%

5 0
3 years ago
The activation energy (E*) for 2N2O ---&gt; 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be
Sedbober [7]

Answer:

Explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature T_1 = 1001 K

Final temperature T_2 = 298 K

Applying the equation of Arrhenius theory.

In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})

where ;

R gas constant = 8.314  J/K/mol

In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})

In \dfrac{k_2}{0.380 }= -70.8655

\dfrac{k_2}{0.380 }= e^{-70.8655}

\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}

{k_2}= 1.67303256 \times 10^{-31} \times {0.380 }

{k_2}= 6.3575 \times 10^{-32}  /M .sec

Half life:

At 1001 K.

t_{1/2} = \dfrac{In_2}{k_1}

t_{1/2} = \dfrac{0.693}{0.38}

t_{1/2} = 1.82368 secc

At 298 K:

t_{1/2} = \dfrac{0.693}{6.3575 \times 10^{-32}}

t_{1/2} =1.09 \times 10^{31} \ sec

3 0
3 years ago
Can someone please help me and ill mark you as brainlest
r-ruslan [8.4K]

Answer:

can you show the question so we can read it

3 0
3 years ago
Read 2 more answers
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