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Citrus2011 [14]
3 years ago
13

What features did you use to classify igneous rocks as extrusive or intrusive ?

Chemistry
2 answers:
umka21 [38]3 years ago
6 0

Answer: Igneous rocks may be simply classified according to their chemical/mineral composition as felsic, intermediate, mafic, and ultramafic, and by texture or grain size: intrusive rocks are course grained (all crystals are visible to the naked eye) while extrusive rocks may be fine-grained (microscopic crystals) or glass.

Explanation: Hope this helped! :)

Sati [7]3 years ago
6 0

Answer:

Igneous rocks may be simply classified according to their chemical/mineral composition as felsic, intermediate, mafic, and ultramafic, and by texture or grain size: intrusive rocks are course grained (all crystals are visible to the naked eye) while extrusive rocks may be fine-grained (microscopic crystals) or glass ( ...

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Mass of metal=0.0291. Volume of gas collected over water=25.67mL. Temperature=24.1C. Atmospheric pressure=754.6mmHg. Note: Metal
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1) 732.1 mmHg; 2) 0.001 014 mol; 3) 0.001 014 mol; 4) 28.7 g/mol; 5) 187 ppt.

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1) <em>Partial pressure of hydrogen</em>

You are collecting the gas over water, so

p_{\text{atm}} = p_{\text{H}_{2}} + p_{\text{H}_{2}\text{O}}

p_{\text{H}_{2}} = p_{\text{atm}} - p_{\text{H}_{2}\text{O}}

p_{\text{atm}} = \text{754.6 mmHg}

At 24.1 °C, p_{\text{H}_{2}\text{O}} = \text{22.5 mmHg}

p_{\text{H}_{2}} = \text{754.6 mmHg} - \text{22.5 mmHg} = \textbf{732.1 mmHg}

===============

2) Moles of H₂

We can use the Ideal Gas Law.

<em>pV = nRT</em>                Divide both sides by <em>RT</em> and switch

<em>n</em> = (<em>pV</em>)/(<em>RT</em>)

p = 732.1 mmHg                                                 Convert to atmospheres

p = 732.1/760                                                      Do the division

p = 0.9633 atm

V = 25.67 mL                                                        Convert to litres

V = 0.025 67 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 24.1 °C                                                              Convert to kelvins

T = (24.1 + 273.15 ) K = 297.25 K                          Insert the values

n = (0.9633 × 0.025 67)/(0.082 06 × 297.25)     Do the multiplications

n = 0.02473/24.39                                                 Do the division

n = 0.001 014 mol

===============

3)<em> Moles of metal </em>

The partial chemical equation is

M + … ⟶ H₂ + …

The molar ratio of M:H₂ is 1 mol M:1 mol H₂.

Moles of M = 0.001 014 × 1/1                          Do the operations

Moles of M = 0.001 014 mol M

===============

4) Atomic mass of M

Atomic mass = mass of M/moles of M     Insert the values

Atomic mass = 0.0291/0.001 014             Do the division

Atomic mass = 28.7 g/mol

===============

5) <em>Relative deviation in ppt </em>

Your metal must be in Group 2 because of the 1:1 molar ratio of M:H₂.

The metal with the closest atomic mass is Mg (24.305 g/mol).

Relative deviation in ppt = |Experimental value – Theoretical value|/Theoretical value × 1000

Relative deviation = |28.7 – 23.405|/23.405 × 1000     Do the subtraction

Relative deviation = |4.39|/23.405 × 1000                      Do the operations

Relative deviation = 187 ppt

3 0
3 years ago
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