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3 years ago

Sorry for asking for too much but me needs helppppp

Chemistry

2 answers:

zalisa [80]3 years ago

7 0

Answer:

physical

Explanation:

pashok25 [27]3 years ago

5 0

Answer:

Physical property

Explanation:

A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties include color, density, hardness, and melting and boiling points. A chemical property describes the ability of a substance to undergo a specific chemical change.

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What is silica sand used for?

ivann1987 [24]

Microchips I know for sure and some types of glass

6 0

4 years ago

Zinc Mass If a 1.85 g mass of zinc produces 475 mL of gas and your balloon weighs 0.580 g and the room temperature is 21.5°C. Ca

Alexus [3.1K]

The weight of the balloon is irrelevant because it is the gas that lifts it in the air. We are already given with the required volume, so we use this instead. The atomic weight of zinc is 65.38 g/mol. Assuming ideal gas behavior,

PV=nRT
P(475 mL)(1 L/1000 mL) = (1.85/65.38)(0.0821 L·atm/mol·K)(21.5 + 273)
P = 1.44 atm

Then, we use this pressure and the volume to find the moles of zinc.

(1.44 atm)(475 mL+1 mL)(1 L/1000 mL) = n(0.0821 L·atm/mol·K)(21.5 + 273)
Solving for n,
n = 0.02836 moles of zinc

3 0

4 years ago

Consider the following reaction: COCl2(g) ⇌ CO(g) + Cl2(g) A reaction mixture initially contains 1.6 M COCl2. Determine the equi

professor190 [17]

Answer:

The equilibrium concentration of CO is 0.0361 M

Explanation:

Step 1: Data given

Kc = 8.33 *10^-4

Molarity of COCl2 = 1.6 M

Step 2: The balanced equation:

COCl2(g) ⇌ CO(g) + Cl2(g)

Step 3: Calculate final concentrations

The initial concentration of COCl2 = 1.6M

The initial concentration of CO and Cl2 = 0M

There will react xM of COCl2

Since the mole ratio is 1:1

The final concentration of CO and Cl2 will be X M

The final concentration of COCl2 will be (1.6 -X)M

Step 4: Define Kc

Kc= [CO] *[Cl2] / [COCl2] = 8.33*10^-4

Kc = X*X / 1.6-X = 8.33 * 10^-4

8.33 * 10^-4 = X² /(1.6-X)

8.33 * 10^-4 *(1.6 -X) = X²

0.0013328 - 8.33*10^-4 X = X²

X² + 8.33*10^-4 X - 0.0013328= 0

X = 0.0361 M = [CO] = [Cl2]

[COCl2] = 1.6 - 0.0361 = 1.5639 M

To control this we can calculate the Kc

(0.0361*0.0361)/1.5639 = 0.000833

5 0

3 years ago

Question 7 options: The cell potential of an electrochemical cell made of an Fe, Fe2 half-cell and a Pb, Pb2 half-cell is _____

larisa [96]

Answer: Thus the cell potential of an electrochemical cell is +0.28 V

Explanation:

The calculation of cell potential is done by :

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Fe^{2+}/Fe]}= -0.41V$

$E^0_{[Pb^{2+}/Pb]}=-0.13V$

As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.

$E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Fe^{2+}/Fe]}$

$E^0=-0.13- (-0.41V)=0.28V$

Thus the cell potential of an electrochemical cell is +0.28 V

5 0

3 years ago

Please somebody give me the answers

RSB [31]

Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams (1 mole of CO2 = 44 gram )

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

$\frac{18}{32}\times 21$

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0

3 years ago