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krok68 [10]
3 years ago
9

Consider the reaction when aqueous solutions of copper(ii) nitrate and nickel(ii) chloride are combined.

Chemistry
1 answer:
mezya [45]3 years ago
8 0
Answer : The reaction that will occur when you combine the aqueous solutions of copper nitrate with nickel chloride is a displacement reaction.

It explained below with the reaction :-

Explanation : Cu(NO _{3}) _{2}  _{(aq)}   + NiCl _{2} _{(aq)}  ----\ \textgreater \  CuCl _{2} _{(s) } + Ni(NO _{3}) _{2}_{(aq)}

The reaction will give the major product as copper chloride in the form of solid. Rest all the ions will  remain in the aqueous form.
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a sample of carbon dioxide is contained in a 250.0mL flask at 0.973 atm and 19.0 celcius. how many molecules of gas are in the s
lbvjy [14]

Answer:n = PV/RT = 0.923 atm x 0.250 L / (0.082057 x 290.75 K) . n = 0.00967. mole  0.00967 mole x 6.22x10^23 molecules/mole = 6.02x10^21 molecules of gas

Explanation:

3 0
3 years ago
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What needs to happen at point 3 for igneous rock to transform into sedimentary rock?
lara [203]
An igneous rock is crushed into little shards of rock called sediment, the sediment is then packed together by other rocks or strong forces, that form together to make a Sedimentary rock.
4 0
2 years ago
Which is the next logical step in balancing the given equation? CS2(1)+Cl2(g) CCl4(1) S2Cl2(1)
Mandarinka [93]
Reaction of current interest is: 

<span>CS2(l)       +          Cl2(g)          </span>→       <span>CCl4(l)     +     S2Cl2(l)

While balancing the chemical reaction, care must be taken that number of atoms of reactant side is equal to number of reactant on product side. 

In present case, There is 1 'C' atom of both reactant and product side
There are 2 'S' atoms on both reactant and product side.
However, there are 2 Cl atoms of reactant side, but 6 Cl atoms on product side.
Hence multiplying Cl2 by 3, would equal the number of Cl atoms on both the sides.

Thus, the balanced reaction is
</span>CS2(l)       +         3 Cl2(g)          →       CCl4(l)     +     S2Cl2(l)
7 0
3 years ago
Air is made up of about 21% oxygen and 78% nitrogen.
gladu [14]
I believe its A "Air is made up of about 21% oxygen and 78% nitrogen. In this solution, oxygen is the solute and nitrogen is the solvent.
6 0
3 years ago
At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initiall
mestny [16]

Answer:

concentration of [O_2] = 0.0124 = 12.4 ×10⁻³ M

concentration of [CO] = 0.0248 = 2.48 ×10⁻² M

concentration of [CO_2] = 0.4442 M

Explanation:

Equation for the reaction:

2CO_2_{(g)                ⇄          2CO_{(g)       +       O_2_{(g)

Concentration of   CO_2_{(g) = \frac{2.3}{4.9}  = 0.469

For our ICE Table; we have:

                       2CO_2_{(g)                ⇄          2CO_{(g)       +       O_2_{(g)

Initial                 0.469                              0                           0

Change              - 2x                                +2x                      +x

Equilibrium       (0.469-2x)                       2x                         x

K = \frac{[CO]^2[O]}{[CO_2]^2}

K = \frac{[2x]^2[x]}{[0.469-2x]^2}

4.1*10^{-6}=\frac{2x^3}{(0.469-2x)^2}

Since the value pf K is very small, only little small of  reactant goes into product; so (0.469-2x)² = (0.469)²

4.1*10^{-6} = \frac{2x^3}{(0.938)}

2x^3 =3.8458*10^{-6

x^3 =\frac{3.8458*10^{-6}}{2}

x^3=1.9229*10^{-6

x=\sqrt[3]{1.9929*10^{-6}}

x = 0.0124

∴ at equilibrium; concentration of  [O_2] = 0.0124 = 12.4 ×10⁻³ M

concentration of [CO] = 2x  = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of [CO_2] = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M

3 0
3 years ago
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