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docker41 [41]
3 years ago
8

Iron is much denser than a feather. yet, a particular sample of feathers weighs more than a sample of iron. explain how this is

possible.
Chemistry
2 answers:
White raven [17]3 years ago
6 0
Answer is: density doesn't affect on weigh.
Density <span>of a substance is its </span>mass<span> per unit </span>volume, <span>density is defined as mass divided by volume (d = m/V).
</span>Weight is force<span> on the object due to </span>gravity. Weight is <span>product of the </span>mass<span> of substance and the magnitude of the local </span>gravitational acceleration (<span>W = <span>mg). Feather can have larger mass and also weight.</span></span>
igomit [66]3 years ago
4 0
<span>How iron can be more dense as a feather yet more feathers can weigh more is because density and weight are two completely different things. No matter how many feathers there are it's density never changes, it's weight is able to change but not density. Therefore the density of iron is higher no matter what but may weigh less than a group of feathers.</span>
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Explanation:

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EXPERIMENT 1: Identify the oxidation and reduction half-reactions that occur in
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Answer:

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Anode: Zn²⁺  +  2e⁻  →  Zn   (Reduction)

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Explanation:

To identify the half reaction we need to see the oxidation states.

Mn(s) → Ground state → Oxidation state = 0

Mn(NO₃)₂ → Mn²⁺  → The oxidation state has increased.

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Mn → Mn²⁺  +  2e⁻

Zn(NO₃)₂ →  Zn²⁺

Zn → Ground state → The oxidation state was decreased.

This is the reduction reaction. It has gained two electrons:

Zn²⁺  +  2e⁻  →  Zn

Cathode: Mn → Mn²⁺  +  2e⁻

Anode: Zn²⁺  +  2e⁻  →  Zn

Mn | Mn²⁺ ||  Zn²⁺  | Zn

4 0
3 years ago
How is ionic radius<br> measured?
worty [1.4K]

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3 years ago
The air in a 4.00 L tank has a pressure of 1.20 atm . What is the final pressure, in atmospheres, when the air is placed in tank
stellarik [79]

Answer:

P₂ = 1.92 atm

Explanation:

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P₁V₁= P₂V₂

Given data:

Initial volume = 4.00 L

Initial pressure = 1.20 atm

Final volume = 2500  mL  

Final pressure = ?

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First of all we will convert the volume into litter.

2500 mL × 1L/1000 mL = 2.5 L

P₁V₁ =  P₂V₂

P₂ = P₁V₁ /V₂

1.20 atm×4.00 L = P₂ ×2.5 L

P₂ = 1.20 atm×4.00 L/ 2.5 L

P₂ = 4.8 atm. L/ 2.5 L

P₂ = 1.92 atm

4 0
4 years ago
A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re
Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of \gamma.

\gamma=\frac{C_p}{C_v}

As, C_p=R+C_v

So, \gamma=\frac{R+C_v}{C_v}

Given :

C_v=20.8J/K\\\\R=8.314J/K

\gamma=\frac{8.314+20.8}{20.8}=1.4

Now we have to calculate the initial volume of gas.

Formula used :

P_1V_1=nRT_1

where,

P_1 = initial pressure of gas = 4.25 atm

V_1 = initial volume of gas = ?

T_1 = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)

V_1=5.80L

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

P_1V_1^{\gamma}=P_2V_2^{\gamma}

where,

P_1 = initial pressure of gas = 4.25 atm

P_2 = final pressure of gas = 2.50 atm

V_1 = initial volume of gas = 5.80 L

V_2 = final volume of gas = ?

\gamma = 1.4

Now put all the given values in above formula, we get:

(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}

V_2=8.47L

Now we have to calculate the final temperature of gas.

Formula used :

P_2V_2=nRT_2

where,

P_2 = final pressure of gas = 2.50 atm

V_2 = final volume of gas = 8.47 L

T_2 = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2

T_2=257.9K\approx 258K

Now we have to calculate the work done.

w=nC_v(T_2-T_1)

where,

w = work done = ?

n = number of moles of gas =1.0 mol

T_1 = initial temperature of gas = 300 K

T_2 = final temperature of gas = 258 K

C_v=20.8J/K

Now put all the given values in above formula, we get:

w=(1.0mol)\times (20.8J/K)\times (258-300)K

w=-873.6J

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0
3 years ago
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