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3 years ago

Q1... list ten metal ..?

Chemistry

2 answers:

Roman55 [17]3 years ago

7 0

Gold.

Silver.

Iron.

Copper.

Nickel.

Aluminum.

Mercury( Liquid metal)

Titanium.

alexandr402 [8]3 years ago

7 0

Answer:

Rhodium.

Platinum.

Gold.

Ruthenium.

Iridium.

Osmium.

Palladium.

Rhenium.

Silver.

Indium.

Explanation:

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What volume of 2.50 M Na2SO4 solution contains 71.0 grams of Na2SO4?

statuscvo [17]

Answer

0.02

Explanation:

7 0

4 years ago

When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction:

slamgirl [31]

Answer:

1.5 mole

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

Step 2:

Determination of the masses of Al and Cl2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar Mass of Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 from the balanced equation = 3 x 71 = 213g

From the balanced equation,

54g of Al reacted.

213g of Cl2 reacted

Step 3:

Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

54g of Al reacted with 213g of Cl2.

Therefore, 40.5g of Al will react with = (40.5 x 213)/54 = 159.75g of Cl2.

From the calculations made above, there are leftover of Cl2 as 159.75g reacted out of 212.7g. Therefore, Cl2 is the excess reactant and Al is the limiting reactant.

Step 4:

Determination of the number of mole in 40.5g of Al. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al = 40.5g

Number of mole of Al =?

Number of mole = Mass/Molar Mass

Number of mole of Al = 40.5/27

Number of mole of Al = 1.5 mole

Step 5:

Determination of the number of mole of AlCl3 produced When 40.5 g of Al and 212.7 g of Cl2 combine together. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

From the balanced equation above,

2 moles of Al produced 2 moles of AlCl3.

Therefore, 1.5 mole of Al will also produce 1.5 mole of AlCl3.

From the calculations made above, 1.5 mole of AlCl3 is produced When 40.5 g of Al and 212.7 g of Cl2 combine together.

8 0

3 years ago

Read 2 more answers

If 21 mL of gas is subjected to a temperature change from 10.0°C to 120°C and a pressure change from 1.0 atm to 15 atm, the new

poizon [28]

The new volume is 1.9 ml

<u><em> calculation</em></u>

The new volume is calculated using the combined gas equation

that is P1V1/T1 = P2V2/T2 where

P1= 1.0 atm

V1= 21 ml

T1 = 10.0 c into kelvin = 10 + 273 = 283 k

P2 = 15 atm

V2=? Ml

T2 = 120 c into kelvin = 120 +273 = 393 K

make V2 the subject of the formula by multiplying both side by T2/P2

V2=T2P1v1 / P2T1

V2=[(393 k x 1.0 atm x 21 ml) / (15 atm x 283 K)] = 1.9 Ml

3 0

3 years ago

A student in chemistry 150-02 weighed out 55.5 g of octane (C8H18) and allowed it to react with oxygen, O2. The products formed

Anni [7]

Answer:

Explanation:

Given data:

Mass of octane = 55.5 g

Balanced chemical equation = ?

Mass of oxygen required to react = ?

Mass of CO₂ for med = ?

Molecules of water produced = ?

Mass of octane required to produced 30.0 g of water = ?

Solution:

Chemical equation:

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Mass of oxygen required to react = ?

Mass of octane = 55.0 g

Solution:

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with oxygen.

C₈H₁₈ : O₂

2 : 25

0.48 : 25/2×0.48 = 6 mol

Mass of oxygen required:

Mass = number of moles × molar mass

Mass = 6 mol × 32 g/mol

Mass = 192 g

Given data:

Mass of carbon dioxide produced = ?

Mass of octane = 55g

Solution:

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with CO₂.

C₈H₁₈ : CO₂

2 : 16

0.48 : 16/2×0.48 = 3.84 mol

Mass of CO₂ produced:

Mass = number of moles × molar mass

Mass = 3.84 mol × 44 g/mol

Mass = 168.96 g

Given data:

Molecules of water produced = ?

Mass of octane = 55g

Solution:

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with H₂O.

C₈H₁₈ : H₂O

2 : 18

0.48 : 18/2×0.48 = 4.32 mol

Number of molecules of water:

1 mol = 6.022× 10²³ molecules

4.32 mol × 6.022× 10²³ molecules/ 1 mol

26 × 10²³ molecules

Given data:

Mass of octane required = ?

Mass of water produced = 30 g

Solution:

Number of moles of water.

Number of moles = mass/ molar mass

Number of moles = 30 g/ 18 gmol

Number of moles = 1.67 mol

Now we will compare the moles of water and octane from balance chemical equation:

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

H₂O : C₈H₁₈

18 : 2

1.67 : 2/18×1.67 = 0.185 mol

Mass of octane:

Mass = number of moles ×molar mass

Mass = 0.185 × 114.23 g/mol

Mass = 21.13 g

5 0

4 years ago

Na + MgF^2 = NaF+ Mg

raketka [301]

Answer:

HAAAAA U IN SOME DEEP S***

Explanation:

5 0

4 years ago

Q1... list ten metal ..?​

Q1... list ten metal ..?