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garri49 [273]
3 years ago
11

A geologist sends you a sample that was collected in the field. you find that it is a gray-colored igneous rock containing amphi

bole and intermediate plagioclase feldspar. using bowens reaction series as a guide, how would you describe its composition?
Chemistry
1 answer:
Bogdan [553]3 years ago
7 0
A grey coloured rock with amphibole and intermediate plagioclase like an andesine would classify as an intermediate rock by Bowen's Reaction Series and by the classification of igneous rocks would probably be like a diorite which is intermediate between a gabbro and a granite. A diorite essentially has no quartz but has the silicates amphibole (like hornblende), mica perhaps a little pyroxene and andesine plagioclase. 
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Masteriza [31]

Answer:

The correct answer is 0, 235 mol

Explanation:

We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol:

1 atm x 5, 25l = n  x 0, 082 l atm / K mol x 273 K

n= 1 atm x 5, 25l /0, 082 l atm / K mol x 273 K

n= 0, 235 mol

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Moles of Zn(NO3)2 in 131.50g of this substance.
Degger [83]
<span>number of moles= mass / molecular mass
mass=131.50g
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3 years ago
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An unknown compound has the following chemical formula: Mg_xCl_2
Art [367]

Answer:

Explanation:

Given parameters:

number of moles magnesium = 6.80mol

number of moles of chlorine = 13.56mol

To find the complete chemical formula, we should obtain the formula of the compound.

                                Mg                            Cl

Number of

moles                      6.8                           13.56

Dividing

by the smallest      6.8/6.8                    13.56/6.8

                                    1                                2

  The formula of the compound is MgCl₂

This is an ionic compound in which Magnesium loses two electrons that would be gained by Cl atoms requiring just an electron each to complete their octet.

3 0
3 years ago
In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very
SVETLANKA909090 [29]

<u>Answer:</u> The final concentration of potassium nitrate is 5.70\times 10^{-6}M

<u>Explanation:</u>

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2          .......(1)

  • <u>Calculating for first dilution:</u>

M_1\text{ and }V_1 are the molarity and volume of the concentrated KNO_3 solution

M_2\text{ and }V_2 are the molarity and volume of diluted KNO_3 solution

We are given:

M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL

Putting values in equation 1, we get:

7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M

  • <u>Calculating for second dilution:</u>

M_2\text{ and }V_2 are the molarity and volume of the concentrated KNO_3 solution

M_3\text{ and }V_3 are the molarity and volume of diluted KNO_3 solution

We are given:

M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL

Putting values in equation 1, we get:

1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M

Hence, the final concentration of potassium nitrate is 5.70\times 10^{-6}M

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