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olganol [36]
3 years ago
11

The astronaut then measures the abundance of magnesium on the new planet, obtaining the following results:

Chemistry
2 answers:
Evgesh-ka [11]3 years ago
8 0
Solution:

Atomic mass = ΣzₓAₓ  zₓ = relative abundance of the isotope
and Aₓ = the atomic mass of the isotope.

Atomic mass = (0.0946x85.91) + (0.07x86.91) + (0.8354x87.91)
= 87.65 amu
natta225 [31]3 years ago
5 0

<u>Answer:</u> The average atomic mass of Strontium is 87.65 u

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomcic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

  • <u>For _{38}^{86}\textrm{Sr} isotope:</u>

Mass of _{38}^{86}\textrm{Sr} isotope = 85.91 u

Percentage abundance of _{38}^{86}\textrm{Sr} isotope = 9.46 %

Fractional abundance of _{38}^{86}\textrm{Sr} isotope = 0.09460

  • <u>For _{38}^{87}\textrm{Sr} isotope:</u>

Mass of _{38}^{87}\textrm{Sr} isotope = 86.91 u

Percentage abundance of _{38}^{87}\textrm{Sr} isotope = 7.00 %

Fractional abundance of _{38}^{87}\textrm{Sr} isotope = 0.0700

  • <u>For _{38}^{88}\textrm{Sr} isotope:</u>

Mass of _{38}^{88}\textrm{Sr} isotope = 87.91 u

Percentage abundance of _{38}^{88}\textrm{Sr} isotope = 83.54 %

Fractional abundance of _{38}^{88}\textrm{Sr} isotope = 0.8354

Putting values in equation 1, we get:

\text{Average atomic mass of Sr}=[(85.91\times 0.0946)+(86.91\times 0.0700)+(87.91\times 0.8354)]

\text{Average atomic mass of Sr}=87.65u

Hence, the average atomic mass of Strontium is 87.65 u

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prisoha [69]
C is the correct answer




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8 0
3 years ago
Perform the following for Part C of this lab:
kaheart [24]

Answer:

a. 0.0110 L

b. 0.0020 L

c. 0.011 mol

d. 5.5 M

e. 0.66 g

f. 33%

Explanation:

There is some info missing. I will use some values to show you the procedure and then you can replace them with your values.

<em>Titrant (NaOH) concentration: 1.0 M</em>

<em>Vinegar volume: 2.0 mL</em>

<em>Initial buret reading (initial NaOH volume): 0.1 mL</em>

<em>Final buret reading (final NaOH volume): 11.1 mL</em>

<em>a. Calculate the volume of NaOH that was added to the vinegar. Convert this volume to liters. Show your work.</em>

The volume of NaOH is the difference between the final and the initial buret reading.

11.1 mL - 0.1 mL = 11.0 mL × (1 L/1000 mL) = 0.0110 L

<em>b. Convert the measured volume of vinegar to liters. Show your work.</em>

2.0 mL × (1 L/1000 mL) = 0.0020 L

<em>c. Calculate the moles of NaOH using the volume and molarity of NaOH. Show your work. moles = molarity x volume</em>

moles = molarity × volume

moles = (1.0 mol/L) × 0.0110 L = 0.011 mol

<em>d. Since the reaction ratio is 1:1, the moles of acetic acid in the vinegar is equal to the moles of NaOH reacted during the titration. Calculate the molarity of the acetic acid in the vinegar. Show your work. molarity = moles / volume</em>

molarity = moles / volume

molarity = 0.011 mol/0.0020 L = 5.5 M

<em>e. Calculate the grams of acetic acid in the vinegar. Show your work. mass = moles x molar mass (g/mol)</em>

mass = moles × molar mass

mass = 0.011 mol × 60.05 g/mol = 0.66 g

<em>f. Assuming that the density of vinegar is very close to 1.0 g/mL, the 2.0 mL sample of vinegar used in the titration should weigh 2.0  g. Use this to calculate the mass % of acetic acid in the vinegar sample. mass % = (mass acetic acid / mass vinegar) * 100%</em>

mass % = (mass acetic acid / mass vinegar) * 100%

mass % = (0.66 g /2.0 g) * 100% = 33%

6 0
3 years ago
No question, just wanted to know how y'all doing. ​
konstantin123 [22]

Answer:

Ello and im good

Explanation:

6 0
3 years ago
You need to produce a buffer solution that has a pH of 5.31. You already have a solution that contains 10. mmol (millimoles) of
Karolina [17]

Answer:

37 mmol of acetate need to add to this solution.

Explanation:

Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-

pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]

Here pH is 5.31, pK_{a} (acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.

Plug in all the values in the above equation:

5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]

or, mmol of CH_{3}COO^{-} = 37

So 37 mmol of acetate need to add to this solution.

3 0
3 years ago
If 0.850 L of a 5.00-Msolution of copper(II) nitrate, Cu(NO3)2, is diluted to a volume of 1.80 L by the addition of water, what
7nadin3 [17]

The molarity of the diluted solution is 2.36M

M₁L₁ = M₂L₂

M₂ = M₁L₁/L₂

M₂ = (5.00 M x 0.850 L)/ 1.80 L

M₂ = 2.36 M Cu(NO₃)₂

<h3>Molarity </h3>

The volume of a material present in a given volume of solution is known as its molarity (M). The number of moles of a solute in a liter of solution is referred to as molarity. The term "molarity" can also refer to a solution's molar concentration. Molarity (M) is the number of moles of a solute per liter of a solution, also referred to as the molar concentration of a solution. The symbols mol/L or, more commonly, M can be used to represent molarity. The word "molar concentration" refers to the amount of a substance per unit volume of solution and is used to describe the concentration of a chemical species, specifically a solute, in a solution. Molarity, amount, and substance concentration are other names for this term.

Learn more about molarity here:

brainly.com/question/8732513

#SPJ4

8 0
2 years ago
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