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zimovet [89]
3 years ago
11

A newly discovered element, X, has two naturally occurring isotopes. 93.5 percent of the sample is an isotope with a mass of 268

.9 u, and 6.5 percent of the sample is an isotope with a mass of 269.9 u. What is the weighted average atomic mass for this element?. . 268.5 u . . 269.0 u . . 269.4 u . . 269.8 u.
Chemistry
2 answers:
Ray Of Light [21]3 years ago
5 0
The weighted average atomic mass (m) of the newly discovered element X is the sum of the products of its isotopes percentage and mass.

                    m = (0.935) x (268.9 u) + (0.065) x (269.9 u) = 268.965 u

Thus, the weight of the atom is approximately 269.0 u.
Levart [38]3 years ago
4 0
To calculate the average mass of the element, we take the summation of the product of the isotope and the percent abundance. In this case, it is 0.935 * 268.9 amu + 0.065* 269.9 amu. This is equal to an average mass of 268.965 amu.
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3 years ago
Predict the products that will be formed by the reaction below. Al(s)+Na2SO4(s)—>
Ostrovityanka [42]

Answer:

Al₂(SO₄)₃ and Na

Explanation:

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3 years ago
What is the pH of a solution made by mixing 30.00 mL 0.10 M HCl with 40.00 mL of 0.10 M KOH? Assume that the volumes of the solu
nirvana33 [79]

Answer:

pH = 12.15

Explanation:

To determine the pH of the HCl and KOH mixture, we need to know that the reaction is  a neutralization type.

HCl  +  KOH  →  H₂O  +  KCl

We need to determine the moles of each compound

M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl

M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH

The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl

HCl  +  KOH  →  H₂O  +  KCl

3 m       4 m                       -

             1 m                      3 m

As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.

1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume

[OH⁻] 1 mmol / 70 mL = 0.014285 M

- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15

3 0
3 years ago
Which of the temperatures below is most likely to be the boiling point of water at 880 torr?view available hint(s)which of the t
Luden [163]
Boiling point<span>  is the </span>temperature<span> at which the vapor pressure of the liquid equals the surrounding pressure.

Above boiling point point, liquid get converted into vapour.

Now, boiling point of water is 100 oC at room pressure. Room pressure is equal to 760 torr. Thus, at 100 oC, vapour pressure of water becomes equal to 760 torr.

Now, if external pressure is increased to 880 torr, more heat is to be supplied so that vapour pressure of water equals 880 torr.

So, at 880 torr, boiling point of water will be more than 100 oC. In present case, most like the boiling point of water is equal to 105 oC.


</span>
6 0
3 years ago
You need to prepare 1 L of the citric acid/citrate buffer. You have chosen to use Method 1 (see lab presentation). Calculate the
prisoha [69]

Answer:

3.11 is the pH of the buffer

Explanation:

The pH of a buffer is obtained using H-H equation:

pH = pKa + log [Conjugate base] / [Weak acid]

<em>Where pH is the pH of the buffer, pKa = -log Ka = 3.14 for the citric buffer and [] could be taken as the moles of each species.</em>

The citric acid,HX (Weak acid), reacts with NaOH to produce sodium citrate, NaX (weak base) and water:

HX + NaOH → H2O + NaX

That means the moles of NaOH added = Moles of sodium citrate produced

And the resulitng moles of HX = Initial moles - Moles NaOH added

<em>Moles HX and NaX:</em>

Moles NaOH = 0.100L * (0.65mol / L) = 0.065 moles NaOH = Moles NaX

Moles HX = 0.300L * (0.45mol / L) = 0.135 moles HX - 0.065 moles NaOH = 0.070 moles HX

Replacing in H-H equation:

pH = 3.14 + log [0.065mol] / [0.070mol]

pH = 3.11 is the pH of the buffer

8 0
2 years ago
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