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-BARSIC- [3]
2 years ago
7

An element consists of 5.85% of an isotope with mass 53.940 u, 91.75% of an isotope with mass 55.9349 u, 2.12% of an isotope wit

h mass 56.9354 u, and 0.28% of an isotope with mass 57.9333 u. Calculate the average atomic mass, and identify the element.
Chemistry
1 answer:
Gemiola [76]2 years ago
4 0

Answer:

The answer to your question is: 55.84 u. The element is Iron

Explanation:

isotope 1    mass 53.94 u        abundance  5.85%

             2   mass 55.9349      abundance 91.75%

            3    mass 56.9354       abundance 2.12%

             4    mass 57.9333      abundance 0.28%

average atomic mass = (53.94 x 0.0585) + (55.9349 x 0.9175) + (56.9354 x 0.0212) + (57.9333 x 0.0028)

average atomic mass = 3.155 + 51.32 + 1.207 + 0.1622

average atomic mass = 55.84 u. The element is Iron

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A. Its temperature will rise continuously until it completely melts

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Which pieces of equipment would be used to measure the density of a solution?
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Answer:

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A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
10.00 mL of 0.10 M NaOH is added to 25.00 mL of 0.10 M HCl during a titration
bekas [8.4K]

Answer:

Explanation:

10 mL = .01 L .

25 mL = .025 mL .

10 mL of .1 M NaOH will contain .01 x .1 = .001 moles

25 mL of .1M HCl will contain .025 x .1 = .0025 moles

acid will neutralise and after neutralisation moles of acid remaining

= .0025 - .001 = .0015 moles .

Total volume = .01 + .025 = .035 L

concentration of remaining HCl = .0015 / .035

Option D is correct.

= .042857 M

= 42.857 x 10⁻³ M .

pH = - log [42.857 x 10⁻³]

= 3 - log 42.857

= 3 - 1.632

= 1.368 .

3 0
3 years ago
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