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nika2105 [10]
1 year ago
12

What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Chemistry
1 answer:
Vadim26 [7]1 year ago
8 0

Answer:

\boxed {\boxed {\sf 333 \ grams}}

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

Q= mc \Delta T

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

  • ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

  • ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

  • Q= 4500.0 J
  • c= 0.4494 J/g°C
  • ΔT = 30.1 °C

Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

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the melting point of scandium fluoride is 1552°c, explain why scandium fluoride has a high melting point.​
Greeley [361]

Answer:

Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.

hope it will help you......

5 0
1 year ago
Read 2 more answers
In the best Lewis structure for the fulminate ion, CNO–, what is the formal charge on the central nitrogen atom?
Andrews [41]

Answer : The formal charge on the central nitrogen atom is, (+1)

Explanation :

Resonance structure : Resonance structure is an alternating method or way of drawing a Lewis-dot structure for a compound.

Resonance structure is defined as any of two or more possible structures of the compound. These structures have the identical geometry but have different arrangements of the paired electrons. Thus, we can say that the resonating structure are just the way of representing the same molecule.

First we have to determine the Lewis-dot structure of CNO^-.

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, CNO^-

As we know that carbon has '4' valence electrons, nitrogen has '5' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in CNO^- = 4 + 5 + 6 + 1= 16

According to Lewis-dot structure, there are 8 number of bonding electrons and 8 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}

For structure 1 :

\text{Formal charge on O}=6-6-\frac{2}{2}=-1

\text{Formal charge on C}=4-2-\frac{6}{2}=-1

\text{Formal charge on N}=5-0-\frac{8}{2}=+1

For structure 2 :

\text{Formal charge on O}=6-4-\frac{4}{2}=0

\text{Formal charge on C}=4-4-\frac{4}{2}=-2

\text{Formal charge on N}=5-0-\frac{8}{2}=+1

For structure 3 :

\text{Formal charge on O}=6-2-\frac{6}{2}=+1

\text{Formal charge on C}=4-6-\frac{2}{2}=-3

\text{Formal charge on N}=5-0-\frac{8}{2}=+1

The best Lewis-dot structure is, structure 1.

Thus, the formal charge on the central nitrogen atom is, (+1)

3 0
1 year ago
A 7.50 liter sealed jar at 18 °c contains 0.125 moles of oxygen and 0.125 moles of nitrogen gas. what is the pressure in the con
devlian [24]
The ideal gas equation is;
PV = nRT; therefore making P the subject we get;
P = nRT/V
The total number of moles is 0.125 + 0.125 = 0.250 moles 
Temperature in kelvin = 273.15 + 18 = 291.15 K
PV = nRT
P = (0.250 × 0.0821 )× 291.15 K ÷ (7.50 L) = 0.796 atm
Thus, the pressure in the container will be 0.796 atm
4 0
1 year ago
Read 2 more answers
1. Concentrated HCl is 11.7M. What is the
SSSSS [86.1K]

Answer:

The concentration is 0,2925M

Explanation:

We use the formula

C initial  x V initial = C final x V final

11,7 M x 25 ml = C final x 1000 ml

C final= (11,7 M x 25 ml)/1000 ml = 0, 2925 M

(This formula applies to liquid solutions)

6 0
2 years ago
A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro
Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on  combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q

Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J

Heat released on  combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows  :

Heat released on  combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q'

Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J

Heat released on  combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0
1 year ago
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