C+ O2 --> CO2 +60kJ <br> is the reaction exothermic or<br> endothermic?

A 10.0 gram sample of Fe contains how many miles of Fe

Dafna1 [17]

Answer:

1.7857 moles

Explanation:

moles=mass/Mr

10/56=1.7857

moles of iron =1.7857

4 0

2 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:

A)

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

B)

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

Please help ASAP!!

joja [24]

Answer:

The pond has more energy because I is so much larger that the Cup of boiling water. Since that mass of the pond is so much larger, It is generating more energy than a boiling cup of water.

4 0

2 years ago

Read 2 more answers

what grade is learning What amount of thermal energy (heat) in joules required to raise the temperature of 25 grams of water fro

blsea [12.9K]

Answer:

10425 J are required

Explanation:

assuming that the water is entirely at liquid state at the beginning , the amount required is

Q= m*c*(T final - T initial)

where

m= mass of water = 25 g

T final = final temperature of water = 100°C

T initial= initial temperature of water = 0°C

c= specific heat capacities of water = 1 cal /g°C= 4.186 J/g°C ( we assume that is constant during the entire temperature range)

Q= heat required

therefore

Q= m*c*(T final - T initial)= 25 g * 4.186 J/g°C * (100°C- 0°C) = 10425 J

thus 10425 J are required

7 0

3 years ago

Suppose that 0.50 grams of barium-131 are administered orally to a patient. Approximately how many milligrams of the barium woul

nalin [4]

Two months later 13.8 milligrams of the barium-131 still be radioactive.

<h3>How is the decay rate of a radioactive substance expressed ? </h3>

It is expressed as:

$A = A_{0} \times (\frac{1}{2})^{t/T}$

where,

A = Amount remaining

A₀ = Initial Amount

t = time

T = Half life

Here

A₀ = 0.50g

t = 2 months = 60 days

T = 11.6 days

Now put the values in above expression we get

$A = A_{0} \times (\frac{1}{2})^{t/T}$

$= 0.50 \times (\frac{1}{2})^{60/11.6}$

$= 0.50 \times (\frac{1}{2})^{5.17}$

= 0.50 × 0.0277

= 0.0138 g

= 13.8 mg [1 mg = 1000 g]

Thus from the above conclusion we can say that Two months later 13.8 milligrams of the barium-131 still be radioactive.

Learn more about the Radioactive here: brainly.com/question/2320811

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Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Suppose that 0.50 grams of ban that 0.50 grams of barium-131 are administered orally to a patient. Approximately many milligrams of the barium would still be radioactive two months later? The half-life of barium-131 is 11.6 days.

3 0

1 year ago

C+ O2 --> CO2 +60kJ is the reaction exothermic or endothermic?