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3 years ago

Ignore thisthanks (:

Chemistry

2 answers:

Vikki [24]3 years ago

6 0

The answer is ignore this thanks

Misha Larkins [42]3 years ago

4 0

Answer:

Really, well I didn't.

SOrry.

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A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to th

Furkat [3]

Answer:

a) The energy transferred is 6.91 kJ

b) The internal energy is 4.90 kJ

c) The work done on the gas is - 2.01 kJ

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 2.00 moles

Pressure = constant

Temperature is heated from 294 K to 414 K

Molar heat capacity of hydrogen gas = 28.8 J/mol*K

Step 2: Calculate the energy transferred to the gas by heat.

Q = n* Cp * ΔT

⇒with Q =the energy transferred

⇒with n = the number of moles = 2.00 moles

⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K

⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

Q = 2.00 * 28.8 * 120

Q = 6912 J = 6.91 kJ

Step 3: Calculate the increase in its internal energy.

ΔEint = n*Cv*ΔT

⇒with ΔEint = the increase in its internal energy.

⇒with n = the number of moles = 2.00 moles

⇒with Cv = The constant volume = 20.4 J/mol*K

⇒with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

ΔEint = 2.00 * 20.4 * 120

ΔEint =4896 J = 4.90 kJ

Step 4: Calculate the work done on the gas.

Work done on the gas = -Q + ΔEint

W = -6.91 kJ + 4.90 kJ

W = -2.01 kJ

6 0

3 years ago

What pressure will be produced when 2.0 moles of N2 gas is heated to 68oC in a container that holds 1.25 of gas?

Stella [2.4K]

The pressure of the nitrogen gas produced is determined as 44.77 atm.

<h3>What is the pressure of the Nitrogen gas?</h3>

The pressure of the nitrogen gas is determined from ideal gas equation, as shown below;

PV = nRT

P = nRT/V

where;

n is number of moles = 2 moles
R is ideal gas constant = 0.08205 L.atm/mol.K
T is temperature = 68⁰C = 68 + 273 = 341 K
V is volume = 1.25 L

P = (2 x 0.08205 x 341)/(1.25)

P = 44.77 atm.

Learn more about pressure here: brainly.com/question/25736513

#SPJ1

3 0

2 years ago

How much volume (in cm^3)is gained by a person who gains 11.8 lb of our fat? Human fat has a density of 0.918 g/cm^3

Zinaida [17]

The volume (in cm³) gained by a person who gains 11.8 lb of fat is 5830.49 cm³

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

<h3>How to convert pounds to grams </h3>

1 lb = 453.592 g

Therefore,

11.8 lb = 11.8 × 453.592

11.8 lb = 5352.3856 g

<h3>How to determine the volume </h3>

Mass = 5352.3856 g
Density = 0.918 g/cm³
Volume =?

Volume = mass / density

Volume = 5352.3856 / 0.918

Volume = 5830.49 cm³

Learn more about density:

brainly.com/question/952755

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3 0

2 years ago

A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat

Lady_Fox [76]

Explanation:

(a) The given data is as follows.

Load applied (P) = 1000 kg

Indentation produced (d) = 2.50 mm

BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$

Now, putting the given values into the above formula as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$ = $\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}$

= $\frac{2000}{9.98}$

= 200

Therefore, the Brinell HArdness is 200.

(b) The given data is as follows.

Brinell Hardness = 300

Load (P) = 500 kg

BHI diameter (D) = 10 mm

Indentation produced (d) = ?

d = $\sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}$

= $\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}$

= 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0

3 years ago

What reaction would take place if hot tugsten was surrounded by air?

Inessa05 [86]

The reaction that would take place if hot tungsten was surrounded by air is that it would react to the oxygen in air to form non conducting tungsten oxide and burns out instantly.

8 0

3 years ago