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Snowcat [4.5K]
3 years ago
7

Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr

essure conditions?
Chemistry
1 answer:
leonid [27]3 years ago
6 0
The law that states the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pressure conditions would be the Charles Law. It <span>is an experimental gas </span>law<span> that describes how gases tend to expand when heated. Hope this answers the question.</span>
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Radioactive cobalt-60 is frequently used in treating cancer. It took 24 years for a 10 gram sample to decay to 0.625 grams. What
-BARSIC- [3]

Answer:

Half life is 6 years.

Explanation:

T½ = In2 / λ

Where λ = decay constant.

But N = No * e^-λt

Where N = final mass after a certain period of time

No = initial mass

T = time

N = 0.625g

No = 10g

t = 24 years

N = No* e^-λt

N / No = e^-λt

λ = -( 1 / t) In N / No (inverse of e is In. Check logarithmic rules)

λ = -(1 / 24) * In (0.625/10)

λ = -0.04167 * In(0.0625)

λ = -0.04167 * (-2.77)

λ = 0.1154

T½ = In2 / λ

T½ = 0.693 / 0.1154

T½ = 6.00 years.

The half life of radioactive cobalt-60 is 6 years

5 0
2 years ago
Two iron balls of different mass are heated to 100°C and dropped in water. If the same amount of heat is lost by the two balls t
Kobotan [32]

Now we know that

Q = mc∆T

Where Q is y energy measured in Joules.

m is the mass measured in grams

c is the specific heat of the substance measured in joule per gram degree Celsius.

∆T is the change in temperature measured in degree Celsius.



Let Q1 be the specific heat of the lighter ball.

c1 be the specific heat of the lighter ball.

m1 be the mass of the lighter ball.

∆T1 be the change in the of the lighter ball.


Let Q2 be the specific heat of the heavier ball.

c2 be the specific heat of the heavier ball.

m2 be the mass of the heavierr ball.

∆T2 be the change in the of the heavier ball.


It has been given that the heat lost, that is Q is the same for both the balls of different mass.Which implies Q1= Q2

Specific heat(c) is the same for both the balls since both are made up of iron. c1=c2


Now heat lost by the lighter ball = heat lost by the heavier ball.

Q1= Q2

m1c1∆T1= m2c2∆T2

Since c1=c2

We get

m1/m2= ∆T2/∆T1

Thus we can say since m2>m1,∆T1> ∆T2.

Now initial temperature of both the balls are 100 degree Celsius.

∆T1 = Final temperature(T1 )-100.

∆T2= Final temperature ( T2)-100

Now since the ∆T1> ∆T2 as arrived from the above equation we can conclude that the final temperature of the ball 1 is greater than that of the ball 2. Since the ball 1 as per our assumption is the lighter ball,the final temperature of the ball which has lighter mass is greater than that of the one having a greater mass.

5 0
3 years ago
Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2
Marizza181 [45]

Answer:

\Delta _rH=-1124.14kJ/mol

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}

In such a way, by using the NIST database, we find that:

\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol

Thus, we plug in the enthalpies of formation to obtain:

\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol

Best regards!

8 0
2 years ago
I want to know which ones are molecular equation, complete ionic equation and net ionic equation
NNADVOKAT [17]

Answer:

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

Explanation:

A molecular equation is a balanced chemical equation which shows the reacting species as molecules rather than as componenet ions in their compounds with subscripts written beside the molecules to indicate the state in which they occur in the chemical reaction.

An ionic equation expresses the reacting species as components ions in a chemical reation. All the ions and molecules reacting are shown.

In a net ionic equation, the ions which remain in the ionic state also known as spectator ions are not written as part of the equation.

From the given attachment;

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

8 0
2 years ago
Pleaseee helpppppppp​
ahrayia [7]

Answer:

Potassium

1s2 2s2 2p6 3s2 3p6 4s1

Explanation:

The atom having only one electron its outermost shell must belong to an element in group one of the periodic table.

Having noted that, we proceed to find out what element in group one that has the atom just described in the question.

That atom must belong to an element in the fourth period. The only group 1 element in the fourth period is potassium.

The electron configuration of potassium is;

1s2 2s2 2p6 3s2 3p6 4s1

6 0
3 years ago
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