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Anestetic [448]
3 years ago
6

The diagram below represents the rock cycle.

Chemistry
1 answer:
spayn [35]3 years ago
4 0

Answer:

Heat and Pressure

Explanation:

You might be interested in
Naturally occurring europium (Eu) consists of two isotopes was a mass of 151 and 153 . Europium-151 has an abundance of 48.03% a
Minchanka [31]

Answer:

151.94 u

Explanation:

The atomic mass of Eu is the <em>weighted average</em> of the atomic masses of its isotopes.

We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its <em>percent abundance</em>).

The atomic masses of Eu-151 and Eu-153 are 150.92 u and 152.92 u, respectively.

Set up a table for easy calculation.

0.4803 × 150.92 u =  72.472 u

0.5197 × 152.92 u  = <u> 79.473 u </u>

                 TOTAL = 151.94  u

6 0
3 years ago
Read 2 more answers
What mass of CO is needed to react completely with55.0 g of Fe2O3(s)+CO(g) yield Fe(s)+CO2(g)?
Katarina [22]

Answer:

28.9 g

Explanation:

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Gather all the information in one place</em> with molar masses above the formulas and masses below them.  

M_{r}:     159.69    28.01

              Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

Mass/g:  55.0

1. Use the molar mass of Fe₂O₃ to calculate the moles of Fe₂O₃.

\text{Moles of Fe$_{2}$O$_{3}$} =\text{55.0 g Fe$_{2}$O$_{3}$} \times \frac{\text{1 mol Fe$_{2}$O$_{3}$}}{\text{159.69 g Fe$_{2}$O$_{3}$}}= \text{0.3444 mol Fe$_{2}$O$_{3}$}

2. Use the molar ratio of CO:Fe₂O₃ to calculate the moles of CO.

\text{Moles of CO} = \text{0.3444 mol Fe$_{2}$O$_{3}$} \times \frac{\text{3 mol CO}}{\text{1 mol Fe$_{2}$O$_{3}$}}= \text{1.033 mol CO}

3.Use the molar mass of CO to calculate the mass of CO.

\text{Mass of CO} = \text{1.033 mol CO}  \times \frac{\text{28.01 g CO} }{\text{1 mol CO}}= \textbf{28.9 g CO}  

3 0
3 years ago
Read 2 more answers
The hydrogen atom is not actually electronegative enough to form bonds to xenon. were the xenon-hydrogen bond to exist, what wou
Marat540 [252]

Answer : The structure of XeH_4 will be square-planar.

Explanation :

In the given molecule XeH_4, 'Xe' is the central atom and 'H' is the terminal atom.

Xenon has 8 valence electrons and hydrogen has 1 valence electron. Therefore, the total number of valence electrons are 8 + 4(1) = 12 electrons.

The number of electrons used in Xe-H bonding = 8 electrons

The remaining electrons which are used as lone pair on central atom (Xe) = 12 - 8 = 4 electrons

There are 4 bonding pairs and 2 lone pairs of electrons, they will be arranged in the octahedral arrangement around the central atom with 2 lone pairs of electrons on central atom. The lone pairs are arranged linearly across the central atom. The resulting structure will be square-planar.

The structure of XeH_4 is shown below.

5 0
3 years ago
Nitrogen monoxide and water react to form ammonia and oxygen, like this: (g)(g)(g)(g) Write the pressure equilibrium constant ex
xxTIMURxx [149]

Answer:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

Explanation:

First, we have to write the balanced chemical equation for the reaction. Nitrogen monoxide (NO) reacts with water (H₂O) to give ammonia (NH₃) and oxygen (O₂), according to the following:

NO(g) + H₂O(g) → NH₃(g) + O₂(g)

To balance the equation, we add the stoichiometric coefficients (4 for NH₃ and NO to balance N atoms, then 6 for H₂O to balance H atoms and then 5 for O₂ to balance O atoms):

4 NO(g) + 6 H₂O(g) → 4 NH₃(g) + 5 O₂(g)

All reactants and products are in the gaseous phase, so the equilibrium constant is expressed in terms of partial pressures (P) and is denoted as Kp. The Kp is expressed as the product of the reaction products (NH₃ and O₃) raised by their stoichiometric coefficients (4 and 5, respectively) divided into the product of the reaction reagents (NO and H₂O) raised by their stoichiometric coefficients (4 and 6, respectively). So, the pressure equilibrium constant expression is written as follows:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

6 0
3 years ago
When balancing a chemical equation, the number of h atoms in 2 ch4 is eight?
Viktor [21]
Yes... that is correct.

CH4 is methane so the coefficent in front of it would double the number of atoms of each element
6 0
3 years ago
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