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denis-greek [22]
3 years ago
11

80cm^3 of oxygen gas diffused through a porous hole in 50 seconds how long wiil it take 120cm^3 of nitrogen (iv) oxide to diffus

e through the same hole at the same conditions ?​

Chemistry
1 answer:
inysia [295]3 years ago
5 0

Answer: it will take 89.93secs

Explanation:Please see attachment for explanation

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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.

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3 years ago
Why do real gases not behave exactly like ideal gases?
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The theory assumes that collisions between gas molecules and the walls of a container are perfectly elastic, gas particles do not have any volume, and there are no repulsive or attractive forces between molecules .

5 0
3 years ago
Read 2 more answers
A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ
NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂   (1)        

b) The precipitate formed is Ni(OH)₂  

 

c) The limiting reactant is:

n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles

n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

n = \frac{2}{1}*0.030 moles = 0.060 moles                  

Hence, the limiting reactant is KOH.  

d) The mass of the precipitate formed is:

n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles

m = n*M = 0.010 moles*92.72 g/mol = 0.927 g  

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M  

C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M

C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M

I hope it helps you!                                                                        

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4 years ago
Why do you think the placebo effect works
timurjin [86]

Answer:

The placebo effect is when an improvement of symptoms is observed, despite using a nonactive treatment. It’s believed to occur due to psychological factors like expectations or classical conditioning.

Research has found that the placebo effect can ease things like pain, fatigue, or depression.

hope this will help u :)

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What group of substances does this substance belong to?
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In comparison see it is very easy in goolge
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