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2 years ago

How many moles of c3h6will be consumed when 4.11 mol of co2 are produced in the following equation 2c3h6 +9o2 - 6co2 + 6H2o

Chemistry

1 answer:

slavikrds [6]2 years ago

7 0

Answer: a.1.37 b.4.11 c.6.00 d.12.3

Explanation:

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dangina [55]

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3 years ago

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The solubility of BaCO3(s) in water at a certain temperature is 4.4 10–5 mol/L. Calculate the value of Ksp for BaCO3(s) at this

azamat

Answer : The value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$ .

Solution : Given,

Solubility of $BaCO_3$ in water = $4.4\times 10^{-5}mole/L$

The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.

The equilibrium equation is,

$BaCO_3\rightleftharpoons Ba^{2+}+CO^{2-}_3$

Initially - 0 0

At equilibrium - s s

The Solubility product will be equal to,

$K_{sp}=[Ba^{2+}][CO^{2-}_3]$

$K_{sp}=s\times s=s^2$

$[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L$

Now put all the given values in this expression, we get the value of solubility constant.

$K_{sp}=(4.4\times 10^{-5}mole/L)^2=19.36\times 10^{-10}mole^2/L^2$

Therefore, the value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$ .

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3 years ago

I need the answer for the first question 1.

sp2606 [1]

metal salt acid hydroyon and why cuz I guessed and I haven't t learned this yet and I need points

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3 years ago

Explain why, at room temperature, CO2 is a gas and CS2 is a liquid. You may use diagrams as part of your explanation.

eduard

Explanation:

A. Hydrogen bonding is present in CS2 but not in CO2.

B. CS2 has greater dipole moment than CO2 and thus the dipole-dipole forces in CS2 are stronger.

C. CS2 partly dissociates to form ions and CO2 does not. Therefore, ion-dipole interactions are present in CS2 but not in CO2.

D. The dispersion forces are greater in CS2 than in CO2.

PLS MARK BRAINLIEST :D

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2 years ago

A sample of aluminum is placed in a 25-ml graduated cylinder containing 10.0 ml of water. the level of the water rises to 18.0 m

OLga [1]

To determine the mass of the sample, first find the volume difference after and before the aluminum was placed, the volume change is equal to the volume of the submerged object, in this case aluminum.

Then knowing volume of aluminum and the density of it, we can solve for the mass.

D = m/v

Dv = m

2.7 g/ml • 8 ml = 21.6 grams.

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3 years ago

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