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4 years ago

How is oxygen separated from other substances nearby?

Chemistry

1 answer:

anastassius [24]4 years ago

7 0

Oxygen can be separated by heating certain oxygen compounds by electrolysis or by liquefying air.

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Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C. Record the data and calculated answers for copper

xxTIMURxx [149]

Answer:

Explanation:Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C. Record the data and calculated answers for copper in the 2 tables below. Accepted values for % error calculations can be found below these 2 tables.

DATA

Copper

Lead

Mass of Metal

3 0

3 years ago

In 200 g of a concentrated solution of 70.4 wt% nitric acid (r = 1.41 g/mL, FW(HNO3) = 63.01 g/mol), how many grams of water are

mars1129 [50]

Answer:

59.2 grams

Explanation:

We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:

$W = 200*(1-0.704)\\W=59.2\ g$

There are 59.2 grams of water in this solution.

5 0

3 years ago

1)state Non-Mendillian's Law of Inheritance.

ioda

Answer:

These questions are not chemistry darling. You have chosen the wrong subject

8 0

3 years ago

48.0 mL of 1.70 M CuCl2(aq) and 57.0 mL of 0.800 M (NH4)2S(aq) are mixed together to give CuS(s) as a precipitate. The other pro

artcher [175]

Answer:

The concentration of the Cu(II) ion is 0.777M

Explanation:

Step 1: Data given

Volume of 1.70 M CuCl2 = 48.0 mL = 0.0480 L

Volume of 0.800 M (NH4)2S = 57.0 mL = 0.0570 L

Step 2: The balanced equation

CuCl2 (aq) + (NH4)2S (aq) → 2 NH4Cl (aq) + CuS (s)

Step 3: Calculate moles CuCl2

moles CuCl2 = 0.0480 L * 1.70 M=0.0816 moles

Step 4: Calculate moles (NH4)2S

moles (NH4)2S = 0.0570 L * 0.800 M = 0.0456 moles

Step 5: Calculate the limiting reactant

The ratio between CuCl2 and (NH4)2S is 1 : 1 so (NH4)2S is the limiting reactant . IT will completely be consumed (0.0456 moles).

CuCl2 is in excess. There will remain 0.0816 - 0.0456 = 0.0360 moles

Step 6: Calculate moles of CuS

For 1 mol CuCl2 we need 1 mol (NH4)2S to produce 2 moles of NH4Cl and 1 mol CuS

For 0.0456 moles we'll produce 0.0456 moles CuS

Step 7: Calculate moles of Cu(II)ion

There remains 0.0360 moles CuCl2.

There will be 0.0456 moles CuS produced

Total moles Cu^2+ = 0.0816 moles

Step 8: Calculate concentration of Cu(II) ion

Concentration = moles / volume

Concentration = 0.0816 moles / (0.048+0.057)

Concentration = 0.777 M

The concentration of the Cu(II) ion is 0.777M

7 0

4 years ago

The equilibrium constant for the reaction 2NH3(g) N2(g) + 3H3(g) is Keq = 230 at 300°C. At equilibrium, what will be the value o

evablogger [386]

Answer:

230

Explanation:

Given equilibrium conditions, if a system is at equilibrium, this means the rate of a forward reaction becomes equal to the rate of a reverse reaction and the concentrations of each species become constant (stop to change).

This is defined by the equilibrium constant which defines the concentrations at equilibrium. For the given reaction, the equilibrium constant can be described as the ratio between the concentrations of products (raised to the power of their coefficients) and the concentrations of reactants (raised to the power of their coefficients):

$K_{eq}=\frac{[N_2][H_2]^3}{[NH_3]^2}$

Q, on the other hand, is a reaction quotient. It is used only to determine where the equilibrium will shift when the system is actually not at equilibrium. The expression or Q is exactly same, but concentrations used would be non-equilibrium concentrations. There are three cases:

$Q$ : in this case, equilibrium shifts towards the formation of products;

$Q=K$ : the system is at equilibrium;

$Q>K$ : equilibrium shifts towards the formation of reactants.

Looking at the context of our problem, since this system is at equilibrium, then Q = K = 230.

6 0

4 years ago