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maria [59]
3 years ago
14

Write the formula for the following compounds:

Chemistry
1 answer:
scoundrel [369]3 years ago
4 0
1) NaF
2)MgS
3)Li2O
4)FeCl3
5)Cu3P2
6)MgL2
7)Fe3p
8)AlN
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Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3
Annette [7]

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving  chemical system, in which a reversible chemical reaction takes place.  When this situation is reached, it is observed that the  concentrations of substances, both reagents and reaction products,  they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations  and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }

You know:

  • Kc= 14.5
  • [H₂]= 0.322 M
  • [CH₃OH] =1.56 M

Replacing:

14.5=\frac{1.56}{[CO]*0.322^{2} }

Solving:

[CO]=\frac{1.56}{14.5*0.322^{2} }

[CO]= 1.04 M

The equilibrium value of [CO] is 1.04 M

8 0
3 years ago
The small intestine shown below contains lymphatic vessels but no capillaries. Which nutrient will be absorbed by this small int
Bumek [7]

Answer:

fatty acids

Explanation:

8 0
3 years ago
Read 2 more answers
Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI, and the rate constant is 9.7×10−6M−
Lady bird [3.3K]

Answer : The molarity after a reaction time of 5.00 days is, 0.109 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 9.7\times 10^{-6}M^{-1}s^{-1}

t = time taken  = 5.00 days

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.110 M

Now put all the given values in above equation, we get:

9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)

[A]=0.109M

Hence, the molarity after a reaction time of 5.00 days is, 0.109 M

8 0
3 years ago
Pa help po thanks! I need the answer ASAP
Korolek [52]

The object and description that matches is Object 2 and T.

Object 1 has no matching description.

<h3>Fish Aquarium Filter </h3>

A Fish aquarium filter is a filter whose function is to clean the water of debris, removes the toxic buildup of ammonia and nitrates, and aerates the water so that fish can in a conducive environment and breathe properly

<h3>Engine Oil Filter</h3>

An engine oil filter is a filter whose function is to filter and remove contaminants that may be present in the engine oil, transmission oil, lubricating oil, or hydraulic oil in order for proper functioning of the engine.

Therefore, the object and description that matches is Object 2 and T.

Object 1 has no matching description.

Learn more about filters and their uses at: brainly.com/question/10719424

6 0
2 years ago
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Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
3 years ago
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