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grigory [225]
2 years ago
10

If 801 J of heat is available, what is the mass in grams of iron (specific heat = 0.45 J/g・°C) that can be heated from 22.5°C to

120.0°C?
Chemistry
1 answer:
inysia [295]2 years ago
3 0

Answer:

The correct answer will be "18.25 g".

Explanation:

The given values are:

Specific heat,

C = 0.45 J/g・°C

Heat involved,

q =  801 J

Temperature,

ΔT = 120.0°C-22.5°C

     = 97.5°C

As we know,

⇒  C = \frac{q}{m \Delta T}

On substituting the given values, we get

⇒  0.45=\frac{801}{m(97.5)}

⇒  m = 18.25 \ g

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The symbol P represents the element phosphorus. This element’s atomic number is 15. How many protons and electrons are in a P–3
Leno4ka [110]
A charge of -3 means it gained 3 electrons meaning electrons would go from 15 to 18 because protons and electrons are the same and you start with 15.
So your answer is 15 protons, 18 electrons
7 0
2 years ago
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A thin sheet of iridium metal that is 3.12 cm by 5.21 cm has a mass of 87.2 g and a thickness of 2.360 mm. What is the density o
never [62]

Answer:

Therefore the density of the sheet of iridium is 22.73 g/cm³.

Explanation:

Given, the dimension of the sheet is 3.12 cm by 5.21 cm.

Mass: The mass of an object can't change with respect to position.

The S.I unit of mass is Kg.

Weight of an object is product of mass of the object and the gravity of that place.

Density: The density of an object is the ratio of mass of the object and volume of the object.

Density =\frac{mass}{volume}

            =\frac{Kg}{m^3}                 [S.I unit of mass= Kg and S.I unit of m³]

Therefore the S.I unit of density = Kg/m³

Therefore the C.G.S unit of density=g/cm³

The area of the sheet is = length × breadth

                                        =(3.12×5.21) cm²

                                       =16.2552 cm²

Again given that the thickness of the sheet  is 2.360 mm =0.2360 cm

Therefore the volume of the sheet is =(16.2552 cm²×0.2360 cm)

                                                             =3.8362272 cm³

Given that the mass of the sheet of iridium is 87.2 g.

Density =\frac{87.2 g}{3.8362272 cm^3}

             =22.73 g/cm³

Therefore the density of the sheet of iridium is 22.73 g/cm³.

5 0
2 years ago
. A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture
postnew [5]

Answer:

A) Mass flow rate of air = 22.892 kmol/hr

B)percentage by mass of oxygen in the product gas = 22.52%

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91

Molar mass of CH4 = 12 + 1(4) = 16 g/mol

We are given the average molecular weight of air = 29 g/mol

Thus;

Average molar mass of air and methane mixture is;

M_avg = (0.09 × 16) + (0.91 × 29)

M_avg = 27.83 g/mol

We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h

Thus;

Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr

Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr

We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.

Thus, for 5% of methane, the air required will be;

2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr

Now, the dilution air needed will be =

43.016 - 22.892 = 20.124 kmol/hr

Total mass flow rate of mixture =

700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr

We are told that air consist of 21 mole% Oxygen (O2).

Molar mass of oxygen = 32

Thus;

Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252

Thus, written in percentage form, we have; 22.52%

So, percentage by mass of oxygen in the product gas = 22.52%

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The answer is D; Mercury-194
All of the others are not when I looked them up

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Answer:

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Explanation:

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3 0
2 years ago
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