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Svet_ta [14]
3 years ago
12

Which of the following is composed mostly of ice?

Chemistry
1 answer:
Zanzabum3 years ago
5 0

Answer: The answer is C- comets!                                                                            

Explanation:  Callisto is composed mainly of rock and water ice, although other ices like ammonia ice and carbon dioxide ice may be present. Water ice occurs at the surface of Callisto so it would be comets.

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PLEASE HELP ME!
harina [27]
The pH scale is used to measure the degree of acidity or alkalinity of a solution. The scale runs from 0 (very acidic solutions can have a negative pH) to 14 (very alkaline solutions can have a pH higher than this), while a neutral liquid such as pure water has a pH of 7. The pH is linked to the concentration of hydrogen ions (H +) in the solution. Diluting an acid or alkali affects the concentration of H +<span> ions in a solution and therefore affects the pH. In this activity, we will investigate how diluting an acid or alkali affects the pH.
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8 0
3 years ago
State Hess' law of constant heat summation.
AveGali [126]

Answer:

-74.6 kj/mol

Explanation:

you can see the answer at the pic

8 0
3 years ago
What product, including stereochemistry, is formed when CH3OCH2CH2C≡CCH2CH(CH3)2 is treated with the following reagent: H2 (exce
Bas_tet [7]

Using hydrogen and Lindlar catalyst the triple bond will be hydrogenated to a double one with a cis conformation.  

4 0
3 years ago
How to determine Km and Vm from a graph
ElenaW [278]

Answer:

From the graph find the maximum velocity and half it i.e. Vmax/2. Draw a horizontal line from this point till you find the point on the graph that corresponds to it and read off the substrate concentration at that point. This will give the value of Km.

5 0
2 years ago
If a temperature increase from 10.0 ∘C to 22.0 ∘C doubles the rate constant for a reaction, what is the value of the activation
maria [59]

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\

\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\

\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\

\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}

E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\

E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}

3 0
3 years ago
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