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3 years ago

11) A sample of rhenium of suspected extrasolar origin (translation: not from this solar

Chemistry

1 answer:

jarptica [38.1K]3 years ago

4 0

Answer:

Atomic number 75 is dedicated to an element named rhenium and has been given Re as its chemical name.

Explanation:

With a really low concentration it is one of the rarest metals that is found in Earth's crust.

Like all other elements rhenium also has certain isotopes among with 185 and 187 are the most stable ones. Hence these two are the ones that are naturally available abundance is 34% and 63% respectively.

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3 years ago

Based on your answer in part B, arrange the five elements in order of their reactivity. Drag the most reactive element to the to

amm1812

Answer: Order of the elements from most reactive to least reactive is:

Mg > Zn > Fe > H > Cu

Explanation: Reactivity of the elements is the tendency of the elements to loose or gain electrons with ease.

As, all the given elements are metals, so the metals which will loose electrons easily will be more reactive. To study the reactivity of metals, we use reactivity series.

The metals which are more reactive are placed on the top of the reactivity series and which are less reactive are placed at the bottom.

Hence, the order of the elements from most reactive to least reactive is:

Mg > Zn > Fe > H > Cu

Reactivity series is attached below.

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3 years ago

Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH

uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

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