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2 years ago

Hii I need a tutor for chemistry

Chemistry

1 answer:

kap26 [50]2 years ago

3 0

Answer:

I had a tutor for chemistry but I think I don't have the number now.

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Can someone best explain why a washing machine represents an open system

Effectus [21]

Open system is a system in which mass or energy can be lost to or gained from the environment. In this case, water and power come from external sources best explains why a washing machine would be considered an open system.

6 0

2 years ago

Read 2 more answers

29. What is E for a system which has the following two steps:

jasenka [17]

Answer:

Zero

Explanation:

Recall that;

E = q + w

Where;

q = heat, w = work done

When heat is absorbed by the system q is positive

When heat is evolved by the system q is negative

When the system does work, w is negative

When work is done on the system w is positive

Step 1

ΔE1= 60 KJ + 40 KJ = 100KJ

Step 2

ΔE2= (-30 KJ) + (-70 KJ) = (-100) KJ

ΔE1 + ΔE2= 100KJ + (-100) KJ = 0KJ

4 0

2 years ago

What is the name of the pores, or holes, on the outer layer of the sponge that take water in?

ivanzaharov [21]

The tiny holes in a sponges outer layer that take in water are called ostia, I believe. Hope it helps

7 0

3 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0

3 years ago

In the electroplating of nickel, 0.200 faraday of electrical charge is passed through a solution of NiSO4. What mass of nickel i

Viefleur [7K]

The mass of nickel, Ni deposited when 0.2 faraday of electrical charge is passed through a solution of NiSO₄ is 5.9 g

7 0

1 year ago

Hii I need a tutor for chemistry￼

Hii I need a tutor for chemistry