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3 years ago

Multiply 17.243 × 0.95 = _____

Chemistry

1 answer:

sp2606 [1]3 years ago

5 0

Answer:16.38085

Explanation:Aline the decimals then start multiplying.

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What is the name for the elevation difference between two adjacent contour lines

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Contour interval. Noted on the map legend.

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3 years ago

An aqueous solution of cesium chloride is prepared by dissolving 52.3 g cesium chloride in 60.0g of water at 25°C. The volume of

LekaFEV [45]

Given data:

Mass of cesium chloride CsCl (m) = 52.3 g

Mass of water (mh2O)= 60.0 g

Volume of the solution (V) = 63.3 ml

Step 1: Calculate the number of moles of CsCl

Molar mass of CsCl = 168.36 g/mol

Mass of CsCl = 52.3 g

# moles = mass/molar mass = 52.3 g/168.36 g.mol-1 = 0.3106 moles

Step 2: Volume of solution in Liters

1000 ml = 1 liter

therefore, 63.3 ml corresponds to : 63.3 ml * 1 L/1000 ml = 0.0633 L

Step 3: Calculate molarity of CsCl

Molarity = moles of CsCl/volume of solution

= 0.3106 moles/0.0633 L = 4.91 moles/L

Molarity = 4.91 M

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3 years ago

Appropriate labels for each axis

azamat

Answer:

the vertical is y and the horizontal is x the one that shows depth is the z axis

Explanation:

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3 years ago

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13. The equation used to find acceleration is a =

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Acceleration represented by the equation = Δv/Δt.

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3 years ago

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

Answer: The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

Explanation:

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:

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

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

For $_{77}^{191}\textrm{Ir}$ isotope:

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

For $_{77}^{193}\textrm{Ir}$ isotope:

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

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3 years ago