How many identified isotopes does samarium have?

What are the two categories of chemical sedimentary rock?

AnnyKZ [126]

Calcite and calcite?

4 0

3 years ago

Measurements show that unknown compound X has the following composition: element mass % carbon 41.0% hydrogen 4.58% oxygen 54.6%

anastassius [24]

Answer:

CHO

Explanation:

Carbon = 41%, Hydrogen = 4.58%, oxygen = 54.6%

Step 1:

Divide through by their respective relative atomic masses

41/ 12, 4.58/1, 54.6/16

3.41 4.58 3.41

Step 2:

Divide by the lowest ratio:

3.41/3.41, 4.58/3.41, 3.41/3.41

1, 1, 1

Hence the empirical formula is CHO

8 0

3 years ago

Change the bond between the two carbon atoms in each molecule to a double or triple bond as needed to complete the structure. If

Georgia [21]

Answer:

Your question is complex, because I think you wrote it wrong.

Although in front of this what I can help you is that the carbons are associated between a single, double or triple union.

This depends on whether they are attached to more or less carbons or hydrogens, the carbons have the possibility of joining 4 radicals, both other carbons and hydrogens.

Simple junctions talks about compound organisms called ALKANS.

The double unions, in organic these compounds are called as ALQUENOS.

And as for the tertiary unions, the organic chemistry names them as ALQUINOS.

These compounds that we write, a simple union, the less energy, the less this union, that is why the triple bond is the one that contains the most energy when breaking or destroying it in a reaction.

Explanation:

In a chemical compound the change of these unions if we modified them we would generate changes even in the classifications naming them as well as different compounds and not only that until they change their properties

6 0

3 years ago

What does the pKa of the conjugate acid of a base have to be to remove a hydrogen from water (pKa = 15.7) so that the ratio of h

Inga [223]

Answer:

The pKa of the conjugate acid is 17.7

Explanation:

If hydrogen is removed from water, the equilibrium concentration of the conjugate acid according to the information given in the question becomes,

Kₐ = [OH⁻]/[H₂O]

$K_a=\frac{100}{1} =100$

Now, we determine the equivalent pKa

pKa = -log[ka]

pKa = -log[100]

pKa = -2

Removal of hydrogen from water is reversible as shown below;

H₂O ⇄ OH⁻ + H⁺

15.7 -2

This reaction is reversible, and the difference in pKa = pKa[H₂O] - pKa[H⁺];

pKa of the conjugate acid = 15.7 - (-2) = 17.7

The pKa of the conjugate acid is 17.7

5 0

3 years ago

The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K$

$V=409.9L$

Therefore, the volume of reactant measured at STP left over is 409.9 L

8 0

3 years ago