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natita [175]
3 years ago
15

At-57 °C and 1 atm, carbon dioxide is in which phase? View Available Hint(s) Phase diagrams for water (Figure 1)and carbon dioxi

de (Figure 2)are given here for your reference O gas O liquid O solid O supercritical fluid solid-liquid equilibrium O liquid-gas equilibriunm O solid-gas equilibrium O solid-liquid-gas equilibrium Submit incorrect; Try Again; 3 attempts remaining On the diagram for carbon dioxide, fnd 1 atm on the pressure axis and-57 degrees on the temperature axis. Find where these values intersect, then read which phases are present. Figure 2 of 2 Part D Phase diagram for CO 73 At 10 C and 2 atm carbon dioxide is in the gas phase. From these conditions, how could the gaseous C02 be converted into liquid CO2? View Available Hint(s) Liquid Solid O O O O Increase the temperature. Decrease the temperature. Increase the pressure. Decroase the prossure. 5.2 Gas -78-57 0 31 Temperature (C) Submit

Chemistry
1 answer:
goblinko [34]3 years ago
4 0

Answer:

Gas

Increase the pressure

Explanation:

Let's refer to the attached phase diagram for CO₂ (not to scale).

<em>At -57 °C and 1 atm, carbon dioxide is in which phase?</em>

If we look at the intersection between -57°C and 1 atm, we can see that CO₂ is in the gas phase.

<em>At 10°C and 2 atm carbon dioxide is in the gas phase. From these conditions, how could the gaseous CO₂ be converted into liquid CO₂?</em>

Since at 10°C and 2 atm carbon dioxide is below the triple point, the only way to convert it into liquid is by increasing the pressure (moving up in the vertical direction).

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The nickel(II) ion is commonly dissolved in solution using nickel(II) nitrate hexahydrate, Ni(NO3)2.6H2O. The nickel(II) ion pre
Juliette [100K]
Ni(OH)₂ ⇄ Ni⁺² + 2 OH⁻
Ksp = [Ni⁺²][OH⁻]²  = S (2S)² = 4S³
where S is molar solubility.
at pH = 10 
[H⁺] = 10⁻¹⁰
[H⁺][OH⁻] = 10⁻¹⁴ 
so [OH⁻] = 10⁻⁴ M
Ksp = S [10⁻⁴ + 2S]²
Ksp is very small so the molar solubility of OH⁻ will be very small
so (10⁻⁴ + 2S) is about 10⁻⁴
so Ksp = S x 10⁻⁸
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8 0
3 years ago
Which of the following is NOT soluble in water?
raketka [301]
The correct answer is E. C6H6 is not soluble in water. This is determined by the polarities of these substances. Benzene is a non-polar substance because it has equal number of electrons sharing a bond thus it does not dissolve in water which is a polar substance.
7 0
3 years ago
Read 2 more answers
C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w
Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>ΔH in this case</h3>

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

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7 0
2 years ago
You titrate 25.0 mL of HCl solution of an unknown concentration with a 0.190 M NaOH solution. Complete neutralization of the HCl
galina1969 [7]

A titration is defined as 'the process of determining the quantity of a substance A by adding measured increments of substance B, the titrant, with which it reacts until exact chemical equivalence is achieved (the equivalence point)'.

The Molarity of the HCl Solution requires the addition of 24.6 mL of titrant is 0.0836 M (to 3 significant figures)

<h3>What is Titrant?</h3>

A reagent solution of precisely known concentration that is added in titration

Given reaction HCl + NaOH ===> NaCl + H_2O  neutralization reaction

Note that the mole ratio is 1:1 meaning that 1 mole HCl reacts with 1 mole NaOH to produce 1 mole NaCl and 1 mole H_2O

Find moles of NaOH used:

18.45 ml x 1 L/1000 ml x 0.085 mol/L = 0.002091 moles NaOH used

This meant that there must be 0.002091 moles of HCl present in the 25.0 mls.

We want to find the molarity (M) which is moles HCl/liter of solution.  We know the moles, and we also know the liters of solution (25.0 mls x 1 L / 1000 mls = 0.025 liters)

M = 0.002091 moles HCl / 0.025 liters = 0.08364 M = 0.0836 M (to 3 significant figures)

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4 0
2 years ago
How much energy is gained when a 10.0g sample of liquid water increases in temperature from 13°C to 18°C?
Ivan

When a 10g sample of liquid water increases in temperature from 13°C to 18°C, then the amount of gained energy is 209 joules.

<h3>How do we calculate gained energy?</h3>

The amount of energy which is gained by any sample will be calculated as:

Q = mcΔT, where

Q = gained energy

m = mass of sample = 10g

c = specific heat of water = 4.18 J/g°C

ΔT = change in temperature = 18 - 13 = 5°C

On putting values we get

Q = (10)(4.18)(5)

Q = 209 Joules

Hence required amount of energy is 209 joules.

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4 0
3 years ago
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