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2 years ago

Plz help. Look at the image shown answer 1.2.3 please

Chemistry

1 answer:

horsena [70]2 years ago

6 0

Re enema ughh ten Hessian

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3 is the youngest later

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3 years ago

Consider the reaction and its equilibrium constant at 25.0 ∘C :

olga55 [171]

Since Qp>Kp , the reaction is not at equilibrium.

<h3>What is the equilibrium constant?</h3>

The equilibrium constant shows the extent to which reactants are converted into products.

Now we have to obtain the Qp as follows;

Qp =[CH3OH]/[CO] [H2]^2

Qp = 0.265/(0.265) (0.265)^2

Qp = 14.2

Now we know that Kp = 6.09×10−3, Since Qp>Kp , the reaction is not at equilibrium.

Learn more about equilibrium constant:brainly.com/question/10038290

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1 year ago

What types of damage can wind cause?

Levart [38]

Answer:

Wind between 39-46 mph can cause branches and limbs to break, make it hard for cars to stay on the road wind between 47-54 can cause lighting difficulties, wind between the 60-78 can cause trees to uproot and damage.

Explanation:

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2 years ago

The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of P

Zolol [24]

Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

Pb: 207.2;
S: 32.06.

To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is $\rm PbS$ . There will be 890 grams of $\rm PbS$ in one kilogram of this rock.

Formula mass of $\rm PbS$ :

$M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}$ .

How many moles of $\rm PbS$ formula units in that 890 grams of $\rm PbS$ ?

$\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol$ .

There's one mole of $\rm Pb$ in each mole of $\rm PbS$ . There are thus $\rm 3.71980\; mol$ of $\rm Pb$ in one kilogram of this rock.

What will be the mass of that $\rm 3.71980\; mol$ of $\rm Pb$ ?

$m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg$ .

In other words, the $\rm PbS$ in 1 kilogram of this rock contains $\rm 0.770743\; kg$ of lead $\rm Pb$ .

How many kilograms of the rock will contain enough $\rm PbS$ to provide 1.5 kilogram of $\rm Pb$ ?

$\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg$ .

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3 years ago

What is the mass of 4.75 mol H2SO4

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M(H2SO4)=n*M=4.75*98=465.5g

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3 years ago