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3 years ago

For each of these terms, give an example and a non-example.

1 answer:

4 0

Given what we know, we can confirm that an example is a situation given that corroborates the information shown, while a non-example is one that does not fall in line with the information provided.

<h3>What are examples of the situations given?</h3>

A number that is a multiple of 10 is 40, since 10 times 4 equals 40.
In order to get a product of 10, we can multiply two and five.
To result in a quotient of 10, we can divide one hundred by ten.

<h3>What are non-examples of the situations given?</h3>

one non-example of a multiple of 10 would be to multiply three and seven.
A non-example of a product of 10 is to multiply the number fifty by twenty-five.
To result in a non-example for a quotient of 10, we can divide the number fifteen by three.

Therefore, given the definition of an example as a situation given that corroborates the information shown, while a non-example is one that does not fall in line with the information provided, we can confirm that the ones listed above are correct.

To learn more about examples visit:

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The rate constant is 0.556 L mol-1 s-1 at some temperature. If the initial concentration of NOBr in the container is 0.32 M, how

MrMuchimi

Answer:

12.96 seconds

Explanation:

Assuming the reaction follows a first order

Rate = K[NOBr] = change in concentration of NOBr/time

K = 0.556 L mol^-1 s^-1

Change in concentration of NOBr = 0.32M - 0.039M = 0.281M

0.281/t = 0.556×0.039

t = 0.281/0.021684 = 12.96 seconds

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4 years ago

: Starting with 0.3500 mol CO(g) and 0.05500 mol COCl2(g) in a 3.050-L flask at 668 K, how many moles of Cl2(g) will be present

Mrac [35]

Answer:

The number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

Explanation:

The reaction is:

CO(g) + Cl₂(g) ⇄ COCl₂(g)

The equilibrium constant of the above reaction is:

K = 1.2x10³

To find the moles of Cl₂ present at equilibrium, let's evaluate the reverse reaction:

COCl₂(g) ⇄ CO(g) + Cl₂(g)

The equilibrium constant for the reverse reaction is:

$K_{r} = \frac{1}{1.2 \cdot 10^{3}} = 8.3 \cdot 10^{-4}$

Now, we need to calculate the concentration of CO and COCl₂:

$C_{CO} = \frac{\eta_{CO}}{V} = \frac{0.3500 moles}{3.050 L} = 0.115 M$

$C_{COCl_{2}} = \frac{\eta_{COCl_{2}}}{V} = \frac{0.05500 moles}{3.050 L} = 0.018 M$

Now, from the reaction we have:

COCl₂(g) ⇄ CO(g) + Cl₂(g)

0.018 - x 0.115+x x

The concentration of Cl₂ is:

$K_{r} = \frac{[CO][Cl_{2}]}{[COCl_{2}]}$

$8.3 \cdot 10^{-4} = \frac{(0.115 + x)(x)}{0.018 - x}$

$8.3 \cdot 10^{-4}*(0.018 - x) - (0.115 + x)(x) = 0$

By solving the above equation for x we have:

x = 1.29x10⁻⁴ M = [Cl₂]

Finally, the number of moles of Cl₂ present at equilibrium is:

$\eta_{Cl_{2}} = C_{Cl_{2}}*V = 1.29 \cdot 10^{-4} mol/L*3.050 L = 3.94 \cdot 10^{-4} moles$

Therefore, the number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

I hope it helps you!

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What is the cathode of a galvanic cell made with magnesium and gold?

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Answer:

The cathode of a spontaneous electrochemical cell is the cell in which metal cations undergo reduction. The electrons are supplied by the anode. Without much thought, we can conclude magnesium is much less favorably reduced than gold.

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How would you calculate the number of nanometers in 8.1 cm

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Because 1cm = 1 • 10^7 nm
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