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zhuklara [117]
3 years ago
8

what would happen to an igneous rock exposed at the earth's surface? A.it would melt B.it would weather and erode C.it would cha

nge into a metamorphic
Chemistry
2 answers:
Alex17521 [72]3 years ago
8 0
Hey there,

Answer : C, It would change into a metamorphic

Hope this helps :D

<em>~Top♥</em>
sasho [114]3 years ago
3 0
It would erode if an igneous rock was at the center of the earth it would erode hope this helps

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NaOH + HCl → NaCl + H2O
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The correct answer is B).

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What did you include in your response? check all that apply.
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2Na+2HCI-H,+2Naci
White raven [17]

Answer:

Single Replacement reaction

Explanation:

Reaction Given:

            2Na + 2HCI ------> H₂ + 2NaCl

Type of reaction = ?

Solution:

Look at the reactants and products of the reaction to know the type of reaction

           2Na + 2HCI ------> H₂ + 2NaCl

In this reaction two reactant combine and give 2 products.

Reactant of the reaction

Na = Sodium

HCl = Hydrochloric acid

Product of the reaction

NaCl = sodium chloride

H₂ = hydrogen gas

So,

That sodium metal react with hydrochloric acid and produce salt and water gas

Type of Reaction:

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5 0
3 years ago
When the temperature of a gas in a rigid container decreases, the particles of the gas move slower and experience fewer collisio
4vir4ik [10]
When the temperature of a gas within a rigid container decreases, the particles on average move more slowly and do not collide with one another or the container walls as often. It is the the ideal gas law that applies in this situation and states that a decrease in the temperature of a gas also results in a decrease in the pressure. Thus, automobile tyres which have a colder gas in the winter sometimes need additional air to provide suffiicient pressure. 
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3 years ago
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Density measurements were conducted on a 22.5oC sample of water which had a theoretical density of 0.997655 g/ml. A volume of 10
Sophie [7]

Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).

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p_{average}=\frac{p_{1}+p_{2}+p_{3}}{3}   \\p_{average}=\frac{(0.998+1.041+1.026)g/mL}{3}\\p_{average}=1.022g/mL

  • (3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:

%error=\frac{|p_{average}-p_{theoretical}|}{p_{theoretical}} *100

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%error=2.44 %

7 0
3 years ago
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