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3 years ago

What is the mass of 0.60 moles of Al?

Chemistry

1 answer:

nikitadnepr [17]3 years ago

5 0

It helps to map out how you will navigate through your unit analysis problem before setting it up.

You are given moles and need grams. What can be used as a conversion factor from moles to grams? Molar mass. We are working with aluminum, so we will need the molar mass of aluminum. My Periodic Table tells me the molar mass of aluminum is approximately 27 g/mol. Now we are ready to set up the unit analysis.

$0.60molAl(\frac{27gAl}{1molAl} )=16.2 gAl$

Moles must go on the bottom so that they cancel. Notice how our number of significant figures is 2, so the answer must round to 16 g Al.

<h3>Answer:</h3>

16 grams

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Give the chemical symbol of an element in the third period (row) of the periodic table with four 3p electrons.

Ann [662]

Answer: The chemical symbol of the element is Sulfur.

Explanation:

The element which is present in third period of the periodic table having four 3p electrons is Sulfur. Sulfur is the 16th element of the periodic table which has 6 valance electrons.

The electronic configuration of this element is: $[Ne]3s^23p^4$

This element is considered as a non-metal because it will accept electrons in order to attain stable electronic configuration.

Hence, the chemical symbol of the element is Sulfur.

7 0

3 years ago

Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M

kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

3 0

3 years ago

What is waste good for?

harkovskaia [24]

A because that honestly just makes the most sense

6 0

2 years ago

Write the balanced chemical equation between H2SO4H2SO4 and KOHKOH in aqueous solution. This is called a neutralization reaction

const2013 [10]

Answer:

0.185M sulfuric acid

Explanation:

Based on the reaction:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

1 mole of sulfuric acid reacts with 2 moles of KOH

Initial moles of H₂SO₄ and KOH are:

H₂SO₄: 0.750L ₓ (0.470mol / L) = 0.3525 moles of H₂SO₄

KOH: 0.700L ₓ (0.240mol / L) = 0.168 moles of KOH

The moles of sulfuric acis that react with KOH are:

0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.

Thus, moles that remain are:

0.3525moles - 0.0840 moles = 0.2685 moles of sulfuric acid remains

As total volume is 0.700L + 0.750L = 1.450L, concentration is:

0.2685mol / 1.450L = 0.185M sulfuric acid

8 0

3 years ago

FOR LondonCreamCakes no sure if its right but im trying to help.. lol dont mind all the tabs

MAVERICK [17]

Answer:

It kinda helps but not really

Thanks for trying anyway doe!

Explanation:

7 0

3 years ago