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3 years ago

15

How would granite be classified as

2 answers:

MariettaO [177]3 years ago

5 0

Answer:

it would be classified as a igneous rock.

Explanation:

vredina [299]3 years ago

4 0

Answer:

Granite is an intrusive igneous rock which means it is cooled slowly deep upper the Earth's crust. It is composed of 25% to 35% quartz and over 50% potassium- and sodium rich feldspars.

Explanation:

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How many coppers of each atom in the products

yarga [219]

Answer:

Explanation:

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3 years ago

Which of the following is an example of a chemical change?

Vsevolod [243]

Answer:

D

Explanation:

It is D because once youve put together the ingredients and baked it you can not go in and taake out every ingredient and put it back to how it was before (Please also try to help me with my question it it about scaled copies, Please and thank you!)

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3 years ago

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What is the molar mass of fluorine gas, F2?

Paladinen [302]

Answer:

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Explanation:

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3 years ago

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Nobody answering my questions what the answer givng brainliest

sergiy2304 [10]

Answer: I think it’s B. Because as you can see from the picture there are layers

Explanation:

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3 years ago

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A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

3 0

3 years ago

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