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mariarad [96]
3 years ago
7

Name 3 ores for metals and which metal they yield?​

Chemistry
1 answer:
Sphinxa [80]3 years ago
7 0
Iron, aluminum, and silicon.
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What would happen to the pressure if the volume were reduced to 0.5 l and the temperature increased to 260 ∘c?
Vsevolod [243]
<span>The ideal gas law. PV=nRT pressure x volume = moles x Faraday's constant x Temp Kelvin (C+273) Original data Pressure 1 atmosphere Volume 1 liter Temp 25C = 298K New data Volume 0.5 liter pressure X Temp 260C = 533K P1v1T1 = P2v2T2 plug and chug. (1)(1)(293) = (x)(0.5)(533) Solve for X, which is the new pressure. </span>
8 0
3 years ago
All of the following reactions can be described as displacement reactions except:____________.
Lady_Fox [76]

Answer:

b

Explanation:

The reaction that is not a displacement reaction from all the options is C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}

In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.

<em>Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.</em>

Correct option = b.

8 0
3 years ago
Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica
Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = d_{ideal} = \dfrac{P\times M}{RT}

here:

P = pressure of the O₂ gas = 310 bar

= 310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂  gas = 32 g/mol

∴

d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}

d_{ideal} = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol²    &

b = 0.0319  L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}

310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}

310V^3 -44.389V^2+1.382V-0.044=0

After solving;

V = 0.1152 L

∴

d_{Van \ der \ Waal} = \dfrac{32}{0.1152}

d_{Van \ der \ Waal} = 277.77  g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0
3 years ago
According to this video, how do scientists use data to study/monitor earthquakes and tsunamis?​
Anon25 [30]

Answer:

ewan ko po

Explanation:

sana makatulong

7 0
3 years ago
PLEASE HELP!!!!!!! (in a few minutes please)
Maslowich

<u>B</u>

Hope this helped you, have a good day bro cya)

6 0
3 years ago
Read 2 more answers
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