answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
LuckyWell
1 month ago
13

During metamorphic processes, increased pressure and temperature can affect the _______ of minerals in rock. Rocks subjected to

very high pressure are typically _______ than others because mineral grains are squeezed together, and the atoms are more closely packed. During metamorphic processes, water facilitates the transfer of ions between and within minerals, which can _______ the rate at which metamorphic reactions take place. The growth of new minerals within a rock during metamorphism has been estimated to be about _______ per million years. _______ metamorphism is commonly associated with convergent plate boundaries, where two plates move toward each other. During contact metamorphism, a large intrusion will contain _______ thermal energy and will cool much more slowly than a small one. Metamorphosed sandstone is known as _______. The metamorphic rock _______, made from metamorphosed shale, was once used to make blackboards for classrooms.
Chemistry
2 answers:
castortr0y [3K]1 month ago
6 0
The factors affecting minerals during metamorphism due to elevated pressure and temperature include stability of minerals in rock. Minerals under intense pressure are generally denser because their atomic structures are compacted. Water promotes ion transfer among and within the minerals, which can enhance the rate of metamorphic reactions. Estimates indicate that new minerals in rocks develop at a rate of about 1 millimeter per million years. Wide-scale (regional) metamorphism is usually linked to converging tectonic plate boundaries. In contrast, contact metamorphism occurs when a large intrusion maintains more thermal energy, cooling at a slower rate compared to smaller intrusions. Sandstone that has undergone metamorphosis is termed quartzite. The metamorphic form of shale is known as slate, which was previously utilized for blackboards.
alisha [2.9K]1 month ago
4 0
1.) stability 2.) denser 3.) increase 4.) 1 millimeter 5.) regional 6.) more 7.) quartzite 8.) slate
You might be interested in
A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press
Tems11 [2777]

Answer: The vapor pressure of naphthalene within the flask remains at 2.906\times 10^{-4} atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)

  • The formula employed to determine the enthalpy change for the reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})

The provided information includes:

\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol

Substituting the values into the previous equation produces:

\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol

  • The formula utilized to compute Gibbs free energy change is of a reaction:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]

The equation for the enthalpy change for the reaction is:

\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})

The given factors include:

\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol

By inserting values from the above equation, we arrive at:

\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol

  • For the calculation of K_1 (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

\Delta G^o=-RT\ln K_1

where,

\Delta G^o = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[273+25]K=298K

K_1 = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}

  • To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})

where,

K_2 = Equilibrium constant at 35°C =?

K_1 = Equilibrium constant at 25°C = 1.14\times 10^{-4}

\Delta H = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = 8.314J/K mol

T_1 = Initial temperature = 25^oC=[273+25]K=298K

T_2 = Final temperature = 35^oC=[273+35]K=308K

By plugging values into the equation above, we obtain:

\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}

  • In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for K_p, which is:

K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for K_2 will equal K_p

p_{C_{10}H_8}=2.906\times 10^{-4}

Consequently, the partial pressure of naphthalene at 35°C is 2.906\times 10^{-4} atm.

3 0
1 month ago
A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.
VMariaS [2998]
Jsjsjsdjdjkdskkeekdks probably answer a
3 0
3 months ago
Read 2 more answers
7.744 Liters of nitrogen are contained in a container. Convert this amount to grams.
lorasvet [2795]

Answer:

9.69g

Explanation:

To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.

1 mole of gas takes up 22.4L at STP.

Thus, X moles of nitrogen gas (N2) will fill 7.744L, meaning

X moles of N2 = 7.744/22.4 = 0.346 moles

Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:

Molar Mass of N2 = 2x14 = 28g/mol

Number of moles N2 = 0.346 moles

Find the mass of N2 =?

Mass = number of moles × molar mass

Mass of N2 = 0.346 × 28

Mass of N2 = 9.69g

Hence, 7.744L of N2 consists of 9.69g of N2

7 0
3 months ago
The nitrogen atom of NH2 would have The nitrogen atom of {\rm NH_2} would have blank electrons around the central nitrogen atom.
Alekssandra [3086]

Answer:

(a) The nitrogen atom in contains NH_2^-8 electrons surrounding the central nitrogen atom.

(b) The nitrogen atom in has NH_4^+8 electrons around the central nitrogen atom.

(c) The nitrogen atom in has NH_38 electrons surrounding the central nitrogen atom.

Explanation:

The Lewis dot structure illustrates the connections between atoms within a molecule, additionally showing unpaired electrons present in the molecule.In this structure, valence electrons are represented as 'dots'.

Now we will determine the number of electrons linked to the central nitrogen atom in the specified molecule.

(a) The identified molecule is,

NH_2^-Recognizing that nitrogen has '5' valence electrons while hydrogen has '1', we ascertain the total valence electrons in

= 5 + 2(1) + 1 = 8

According to the Lewis dot structure, this reveals 4 bonding and 4 non-bonding electrons.

NH_2^-From the Lewis structure, we conclude that the nitrogen atom in

has 8 electrons surrounding it.

(b) The identified molecule is, NH_2^-

Knowing that nitrogen has '5' valence electrons and hydrogen has '1' valence electron, the total number of valence electrons for NH_4^+ = 5 + 4(1) - 1 = 8

In the Lewis dot structure, there are 8 bonding electrons and 0 non-bonding electrons.

This leads us to conclude that the nitrogen atom in

contains 8 electrons surrounding it.NH_4^+

(c) The identified molecule is,

NH_4^+

Recognizing nitrogen’s '5' valence electrons and hydrogen’s '1' brings the total valence electrons for

= 5 + 3(1) = 8NH_3Utilizing the Lewis dot structure indicates there are 6 bonding and 2 non-bonding electrons.

According to the Lewis dot structure, we conclude that the nitrogen atom in

has 8 electrons surrounding it.

NH_2^-

3 0
1 month ago
The value of delta G at 141.0 degrees celsius for the formation of phosphorous trichloride from its constituent elements,
VMariaS [2998]
The appropriate answer is option E. Gibbs free energy can be expressed using the equation: ΔG = ΔH - TΔS, where ΔH denotes the change in enthalpy of the reaction, T is the reaction temperature, and ΔS signifies entropy change. For our calculations, we have ΔH = -720.5 kJ/mol which converts to -720500 J/mol (given that 1 kJ = 1000 J), ΔS = -263.7 J/K, and T = 141.0°C, which equals 414.15 K. Consequently, the Gibbs free energy for the specified reaction at 141.0°C is calculated as -611.3 kJ/mol.
6 0
2 months ago
Other questions:
  • A compound has a mass percentage of 53.46% C, 6.98% H, and 39.56% O. What is the empirical formula for this compound
    13·1 answer
  • A solution of cough syrup contains 5.00% active ingredient by volume. If the total volume of the bottle is 11.0 mL , how many mi
    9·2 answers
  • Classify these compounds as acid, base, salt, or other.? 1. ch3oh 3. hno3 5. nabr
    8·2 answers
  • In the quantum mechanical treatment of the hydrogen atom, which one of the following combinations of quantum numbers is not allo
    8·1 answer
  • A mixture of KCl and KNO3
    7·2 answers
  • Part B: Copper (II) chloride (CuCl2; 0.98g) was dissolved in water and a piece of aluminum wire (Al; 0.56g) was placed in the so
    13·1 answer
  • If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series em
    12·1 answer
  • A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a pist
    12·1 answer
  • What is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at 850 mm Hg?
    6·1 answer
  • A buffer is prepared by adding 100 mL of 0.50 M sodium hydroxide to 100 mL of 0.75 M propanoic acid. Is this a buffer solution,
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!