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dolphi86 [110]
4 years ago
6

The composition of rocks affects their densities and the densities of Earth’s layers. Use what you have learned and the informat

ion in the table to answer the question. In which order would these rocks and minerals be found in the layers of Earth from the crust to the lower mantle? eclogite, perovskite, rhyolite, stishovite stishovite, rhyolite, perovskite, eclogite stishovite, perovskite, eclogite, rhyolite rhyolite, eclogite, perovskite, stishovite
Chemistry
2 answers:
Nana76 [90]4 years ago
8 0

Answer:

d

Explanation:

MA_775_DIABLO [31]4 years ago
5 0

Answer:

<u><em>D. rhyolite, eclogite, perovskite, stishovite</em></u>

Explanation:

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The periodic table of elements puts all the known elements into groups with similar properties. This makes it an important tool for chemists, nanotechnologists, and other scientists. If you get to understand the periodic table and learn to use it, you'll be able to predict how chemicals will behave.

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What is the change in vapor pressure when 74.60 g fructose, C6H12O6, are added to 185.5 g water (H2O) at 298 K (vapor pressure o
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Vapor pressure of pure water: 3.1690 kPa

Explanation:

Vapor pressure is P=0.955(3.1690)=3.0263 kPa

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Using noble gas notation write the electron configuration for the silicon atom
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The closest noble gas to silicon is Neon (Ne)

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2 years ago
Kinda need help with these questions
padilas [110]

11. a. Amphoteric

12. a. NH₄⁺, H₂O

Explanation:

11. A substance that can act as either an acid or a base is said to be <u>amphoteric</u>.

Water is an amphoteric substance because it can reaction with either and acid or a base.

HCl + H₂O → Cl⁻ + H₃O⁺

NH₃ + H₂O → NH₄⁺ + OH⁻

12. What are the acids in the following equilibrium reaction?

NH₃ (aq) + H₂O (l) → NH₄⁺ (aq)  +OH⁻ (aq)

H₂O water is acid 1

NH₄⁺ ammonium ion is acid 2

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amphoteric substances

brainly.com/question/1165259

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6 0
3 years ago
Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti
ella [17]

Answer :

The mass of excess mass of Na_2CO_3, AgNO_3,Ag_2CO_3\text{ and }NaNO_3 are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.

Explanation : Given,

Mass of Na_2CO_3 = 4.25 g

Mass of AgNO_3 = 7.50 g

Molar mass of Na_2CO_3 = 106 g/mole

Molar mass of AgNO_3 = 170 g/mole

Molar mass of Ag_2CO_3 = 276 g/mole

Molar mass of NaNO_3 = 85 g/mole

First we have to calculate the moles of Na_2CO_3 and AgNO_3.

\text{Moles of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar mass of }Na_2CO_3}=\frac{4.25g}{106g/mole}=0.040moles

\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{7.50g}{170g/mole}=0.044moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Na_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2NaNO_3

From the balanced reaction we conclude that

As, 2 moles of AgNO_3 react with 1 mole of Na_2CO_3

So, 0.044 moles of AgNO_3 react with \frac{0.044}{2}=0.022 moles of Na_2CO_3

From this we conclude that, Na_2CO_3 is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

The excess mole of Na_2CO_3 = 0.040 - 0.022 = 0.018 mole

Now we have to calculate the mass of excess mole of Na_2CO_3.

\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3=(0.018mole)\times (106g/mole)=1.908g

Now we have to calculate the moles of Ag_2CO_3.

As, 1 moles of AgNO_3 react to give 1 moles of Ag_2CO_3

So, 0.044 moles of AgNO_3 react to give 0.044 moles of Ag_2CO_3

Now we have to calculate the mass of AgCO_3.

\text{Mass of }Ag_2CO_3=\text{Moles of }Ag_2CO_3\times \text{Molar mass of }Ag_2CO_3=(0.044mole)\times (276g/mole)=12.144g

Now we have to calculate the moles of NaNO_3.

As, 2 moles of AgNO_3 react to give 2 moles of NaNO_3

So, 0.044 moles of AgNO_3 react to give 0.044 moles of NaNO_3

Now we have to calculate the mass of NaNO_3.

\text{Mass of }NaNO_3=\text{Moles of }NaNO_3\times \text{Molar mass of }NaNO_3=(0.044mole)\times (85g/mole)=3.74g

6 0
4 years ago
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