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creativ13 [48]
2 years ago
13

A student carried heated a 25.00 g piece of aluminum to a temperature of 100°C, and placed it in 100.00 g of water, initially at

a temperature of 10.0°C. Determine the final temperature of the system (aluminum and water). Useful data: The specific heat of water is 4.18 J/(g°C) The specific heat of aluminum is 0.900 J/(g°C)
Chemistry
1 answer:
SashulF [63]2 years ago
4 0

Answer:

The final temperature of the system is 14.6 °C

Explanation:

<u>Step 1:</u> Data given

mass of the aluminium = 25.00 grams

mass of the water = 100.00 grams

Initial temperature of aluminium = 100 °C

Initial temperature of water = 10.0 °C

Specific heat of water = 4.18 J/g°C

Specific heat of aluminium 0.900 J/g°C

<u>Step 2:</u> Heat transfer

Loss of Heat of the Metal = Gain of Heat by the Water

Qmetal = -Qwater

m(aluminium) * C(aluminium) * ΔT(aluminium) = - m(water) * C(water) * ΔT(water)

25*0.900*(T2-100) = - 100*4.18 * ( T2 - 10)

2250 - 22.5T2 = 418T2 - 4180

T2 =14.6 °C

The final temperature of the system is 14.6 °C

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Jobisdone [24]

Answer:

It is required answer.

Explanation:

Given that :

1. using balanced chemical equation:

ammonium acetate:

The balanced equation is:

NH₃ + H₂O ===> NH₄OH

when ammonia gas dissolves in water then we get the base in form of ammonium hydroxide.

When  NH₄OH reacts with CH₃COOH then we get ammonium acetate and water

NH₄OH + CH₃COOH ===> [CH₃COO]- & NH₄+ & H₂O

So, we can say that,

when we are adding an acid and a base together then we get the product of H₂O and given elements.

2. addition of barium hydroxide to sulfuric acid:

the balanced equation is

H₂SO4+ Ba(OH)₂--> BaSO₄+ 2H₂O

when acid and base reacts together than we get barium sulphate and water

when sulfuric acid and barium hydroxide.

Hence, it is required answer.

8 0
2 years ago
Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at
asambeis [7]

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na +Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g(mass given of NaCl)/58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn,PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres

6 0
2 years ago
A gas sample enclosed in a rigid metal container at room temperature (20.0∘C∘C) has an absolute pressure p1p1p_1. The container
choli [55]

Answer: p2 = 1.06p1

Explanation: pressure increases with temperature increase.

According to Gass law

P1/T1 = P2/T2

T1 = 20°c = 20 +273 = 293k

T2 = 40°c = 40 +373 = 313k

Therefore

P2 = P1T2/T1 = 313P2/293

P2 = 1.06P1

3 0
2 years ago
How many moles of Cobalt are in 27.4 grams of cobalt
xeze [42]
0.4649331785818406 is what 27.4 grams is converted to! You're welcome!! :) 
5 0
2 years ago
Read 2 more answers
In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par
Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0
2 years ago
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