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ser-zykov [4K]
3 years ago
11

What are three terms used to describe organisms such as cow

Chemistry
1 answer:
Zigmanuir [339]3 years ago
7 0
Primary Consumers
Herbivores
Trophic level II Members
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What kinds of things could a magnet make another object or magnet do without touching it?​
Sidana [21]

Answer:

Magnets can be used to make other magnets and things made of iron move without being touched. Something that has been electrically charged can make other things move without touching them.

Explanation:

(can someone help me with my question?)

8 0
3 years ago
Which of the following has higher ionization energy - Be or Ca. Explain your choice in terms of attractive force between protons
Vlad [161]

Answer:

Ca has the greater Ionization Energy because the Trend is I/e decreases as you move down a group. Therefore, Ca has the greater I/e

Explanation:

3 0
3 years ago
Which of the following statements best describes the current atomic theory
blagie [28]
You did not provide possible answers, but one possible might be that the current atomic theory is so sound and plausible that there should not be anything that could change it in the near future.
3 0
3 years ago
Which of the following represent a mixture?
7nadin3 [17]
The answer is B NaCI solid
6 0
3 years ago
The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.
Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]

Where,

\Delta H_v is the Heat of vaoprization (J/mol)

T_b is the normal boiling point of the gas (K)

T_c is the Critical temperature of the gas (K)

P_c is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

T_b=307.4\ K

T_c=466.7\ K

P_c=36.4\ bar

Applying the above equation to find heat of vaporization as:

\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]

\Delta H_v=26400 J/mol

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,

\Delta H_v=26.4 kJ/mol

<u>Option C is correct</u>

6 0
4 years ago
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