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2 years ago

1. If I have 45 L of He in a balloon at 25 degrees celsius and increase the temperature of the

1 answer:

7 0

Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).

We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.

Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.

(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).

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What is the mass number of an ion with 107 electrons, 159 neutrons, and a +1 charge? express your answer as an integer?

Darya [45]

The mass number of an ion is calculated by adding the neutrons and electrons together. In this case 159+107=266. However, since this ion has a +1 charge, you must add 1 to the electrons. This would give a total mass of 267.

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2 years ago

state two methods that can be used to prepare chlorine from rock salt.write an equation in each case

Levart [38]

Answer:

electrolysis of brine

Explanation:

Rock salt deposits are usually mined; occasionally water is pumped down, and brine which contain 25 percent of sodium chloride is found

so d brine is electrolyzed to produce chlorine

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1 year ago

If 3.8 moles of zinc metal react with 6.5 moles of silver nitrate, how many moles of silver metal can be formed, and how many mo

bulgar [2K]

Answer: 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

Explanation:

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = $\frac{1}{2}\times 6.5=3.25moles$ of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal, hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = $\frac{2}{2}\times 6.5=6.5moles$ of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

7 0

2 years ago

Use your knowledge of valence electrons and how they affect bonds to figure out how carbon (Group 14) and oxygen (Group 16)

kenny6666 [7]

Answer: D

Explanation:

O=C=O

5 0

2 years ago

Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca

lina2011 [118]

Answer: The above reaction is non-spontaneous.

Explanation:

For the given chemical reaction:

$Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)$

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.23-0.77=-1.0V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

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2 years ago