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3 years ago

10

What is 10/4 written in a mixed numberd vcccccccccccccccccccccccccccccccc rrrrrrrrrrrrrrrrr xcv,.er 5 ;puiokd nhyh bb uu tr trif

trip tggb guo ric?

1 answer:

Nana76 [90]3 years ago

6 0

10/4 written in a mixed number is 2 1/2

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faltersainse [42]

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3 years ago

A sample of 23.2 g of nitrogen gas is reacted with

slavikrds [6]

Answer:

1.66 moles.

Explanation:

We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.

This is illustrated below:

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 = 23.2 g

Mole of N2 =.?

Mole = mass /Molar mass

Mole of N2 = 23.2/28

Mole of N2 = 0.83 mole

Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.

This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 = 23.2 g

Mole of H2 =?

Mole = mass /Molar mass

Mole of H2 = 23.2/2

Mole of H2 = 11.6 moles

Next, the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

5 0

3 years ago

NNADVOKAT [17]

When you are collecting DNA, you could be looking for a few different things. A few examples could be skin cells, strands of hair, or possibly even a fingernail. Anything that comes from a person, including blood or saliva can be potential DNA that could help investigators to link a person back to a crime.

Investigators do not need a warrant for analyzing crime scenes due to the fact of the dangers of the fire. You must work quickly because accelerants tend to evaporate within days, sometimes hours. It is also important to note that finding the origin of the fire is very important, to make sure it won't be reignited. Debris is usually cleaned away quickly to ensure health and safety issues.

The point of origin of a fire is the lowest point, since fire burns upwards.

High explosive: Ignite almost instantly, like dynamite and TNT. Two different types are primary and secondary.

<em>Primary: easily ignited, very sensitive to heat and friction. often used to ignite other explosives. </em>

<em>Secondary: much less sensitive to heat and friction, might be ignited using other explosive materials. TNT and dynamite are both secondary. </em>

Low explosive: decompose slowly and include black and smokeless powder. They are the most common type of explosives, and are readily available.

4 0

3 years ago

How many orbitals are contained in the third main energy level and in the fifth energy level

antoniya [11.8K]

There are 9 orbitals in the third energy level and 25 orbitals in the fifth energy level.

I hope this helps you.

5 0

3 years ago

An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.

Tju [1.3M]

Answer:

a) $\Delta S$

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

$\delta S=\frac{\delta Q}{T}$

$\Delta S=\frac{Q}{T}$

The energy balance:

$\delta U=[tex]\delta Q- \delta W$

If the process is isothermical the U doesn't change:

$0=[tex]\delta Q- \delta W$

$\delta Q= \delta W$

$Q= W$

The work:

$W=\int_{V1}^{V2}P*dV$

If it is an ideal gas:

$P=\frac{n*R*T}{V}$

$W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV$

Solving:

$W=n*R*T*ln(V2/V1)$

Replacing:

$\Delta S=\frac{n*R*T*ln(V2/V1)}{T}$

$\Delta S=n*R*ln(V2/V1)}$

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

$\Delta S$

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0

3 years ago