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4 years ago

Given the same conditions of temperature and pressure, which noble gas will diffuse most rapidly

2 answers:

0 0

Given the same conditions of temperature and pressure the noble gas He will diffuse most rapidly. He is the smallest noble gas and so it can diffuse (pass through) membranes the easiest.It is colorless, odorless, tasteless, non-toxic chemical element.
Correct answer:A

Natasha2012 [34]4 years ago

0 0

At the same conditions of temperature and pressure the noble gas with the smallest density will diffuse faster. But since density is proportional to the molecular mass. A look at the periodic table shows that Helium will diffuse faster. The correct answer is A.

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In a chemical reaction, sulfur (S8) combines with copper to give a pure compound. If you start with 1.000 g of sulfur (S8) and o

Airida [17]

Answer: The empirical formula for the given compound is $Cu_2S$

Explanation:

We are given:

Mass of pure compound containing copper and sulfur = 4.963 g

Mass of S = 1.000 g

Mass of Cu = (4.963 - 1.000) g = 3.963 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Copper = $\frac{\text{Given mass of Copper}}{\text{Molar mass of Copper}}=\frac{3.963g}{63.55g/mole}=0.0624moles$

Moles of Sulfur = $\frac{\text{Given mass of Sulfur}}{\text{Molar mass of Sulfur}}=\frac{1.000g}{32g/mole}=0.0312moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0312 moles.

For Copper = $\frac{0.0624}{0.0312}=2$

For Sulfur = $\frac{0.0312}{0.0312}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Cu : S = 2 : 1

Hence, the empirical formula for the given compound is $Cu_2S$

8 0

3 years ago

What type of extinguisher will extinguish a class a, b or c fire?

RUDIKE [14]

Any fire extinguisher should put out a fire, that is unless the extinguisher is old and faulty.

7 0

4 years ago

Number of atoms in 5.40gB

Stolb23 [73]

First, take the molar mass of B (10.81g) and divide the value you've been given by the molar mass to get moles (5.4 / 10.81). So you have .5 moles of B and then you need to convert that to atoms. To do that you must use the constant 6.022X10^23 atoms per mole. Take your moles and multiply it by that constant. Your final answer should be (3.01X10^23) atoms.

Hope this helps!

3 0

4 years ago

The steps to balancing a neutralization reaction when given a verbal description can be summarized as follows: Write the chemica

HACTEHA [7]

Answer:

NaHCO3 + HCl ——-> NaCl + H2O + CO2

Explanation:

A neutralization reaction is a chemical reaction between an alkali and an acid to give salt and water as the product.

In the case of carbonates and bicarbonates, an additional product is added. This additional product is carbon iv oxide.

Hence a neutralization reaction involving an acid and a carbonate or bicarbonate would yield water, carbon iv oxide and a salt as the product.

When brioschi reacts with hydrochloric acid, the products are sodium chloride, water and carbon iv oxide.

The equation of the reaction is shown below:

NaHCO3 + HCl ——> NaCl + H2O + CO2

7 0

3 years ago

What is oxidized in a galvanic cell with aluminum and gold electrodes ?

k0ka [10]

Answer:

Aluminum metal

Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

$Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V$

$Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive): $Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Notice that the overall cell potential upon summing is:

$E_{cell}=1.66 V + 1.50 V=3.16 V$

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

6 0

3 years ago