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3 years ago

What provides clues about how igneous rocks were formed? A color B Texture C Luster D Streaks

Chemistry

2 answers:

Troyanec [42]3 years ago

8 0

Did you ever find the answerr? if you did can I pls know ,Ty.

Pani-rosa [81]3 years ago

7 0

the correct answer is C. Texture

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When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago

A 250 cm^3 solution containing 1,46 g of sodium chloride is added to an excess of silver nitrate solution. The reaction is given

faust18 [17]

Answer:

The mass of the precipitate that AgCl is 3.5803 g.

Explanation:

a) To calculate the molarity of solution, we use the equation:

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

We are given:

Mass of solute (NaCl) = 1.46 g

Molar mass of sulfuric acid = 58.5 g/mol

Volume of solution = $250 cm^3 =250 mL$

$1 cm^3= 1 ml$

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{1.46g\times 1000}{58.5g/mol\times 250}\\\\\text{Molarity of solution}=0.09982 M$

0.09982 M is the concentration of the sodium chloride solution.

b) $NaCl (aq)+AgNO_3 (aq)\rightarrow AgCI(s)+NaNO_3(aq)$

Moles of NaCl = $\frac{1.46 g}{58.5 g/mol}=0.02495 mol$

according to reaction 1 mol of NaCl gives 1 mol of AgCl.

Then 0.02495 moles of NaCl will give:

$\frac{1}{1}\times 0.02495 mol=0.02495 mol$ of AgCl

Mass of 0.02495 moles of AgCl:

$0.02495 mol\times 143.5 g/mol=3.5803 g$

The mass of the precipitate that AgCl is 3.5803 g.

3 0

4 years ago

Please help me on this 3 question I need help.

morpeh [17]

It is harm full to a human because it can make a person really sick and it can hurt a plant by killing it

8 0

3 years ago

A chef fills a 50ml container with 43.5g of cooking oil. whats the density of the oil

jasenka [17]

I believe the answer is .87g/mL. But I'm not sure so whatevs

4 0

4 years ago

When Z-4,5-dimethyloct-4-ene is treated with hydrogen chloride, HCl, the result is:________.

Vika [28.1K]

Answer:

The correct option is c

Explanation:

The chemical equation for the reaction of Z-4,5-dimethyloct-4-ene and HCl is shown on the first uploaded image

Now looking at the product we see that there are two who has four different groups attached to them this carbon are known as chiral carbons hence the product formed is a pair of diastereomers

6 0

4 years ago