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valentina_108 [34]

3 years ago

13

A vat of nitrogen at its boiling point (-196 degrees C) absorbs 384000 J of heat. How much mass of nitrogen burns off? (Unit = k

g)

1 answer:

masya89 [10]3 years ago

3 0

Answer:

1.92 kg of nitrogen.

Explanation:

The following data were obtained from the question:

Heat absorbed (Q) = 384000 J

Note: Heat of vaporisation (ΔHv) of nitrogen = 5600 J/mol

Next, we shall determine the number of mole of nitrogen that absorbed 384000 J.

This is illustrated below:

Q = mol·ΔHv

384000 = mole of N2 x 5600

Divide both side by 5600

Mole of N2 = 384000/5600

Mole of N2 = 68.57 moles

Next, we shall convert 68.57 moles of nitrogen, N2 to grams.

This can be obtained as follow:

Molar mass of N2 = 2 x 14 = 28 g/mol.

Mole of N2 = 68.57 moles.

Mass of N2 =..?

Mole = mass /molar mass

68.57 = mass of N2 /28

Cross multiply

Mass of N2 = 68.57 x 28

Mass of N2 = 1919.96 g

Finally, we shall convert 1919.96 g to kilograms.

This can be achieved as shown below:

1000g = 1 kg

Therefore,

1919.96 g = 1919.96/1000 = 1.92 kg.

Therefore, 1.92 kg of nitrogen were burned off.

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What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface

NARA [144]

Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

$g=\frac{GM}{(R+h)^2}$

where:

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

$R=6.37\cdot 10^6 m$ is the Earth's radius

Here,

$h=6.38\cdot 10^6 m$

So the acceleration due to gravity is

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2$

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

$F=mg=(70)(2.45)=171.5 N$

5 0

3 years ago

Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s

Zielflug [23.3K]

Answer:

there will be collision

Explanation:

$v_{s}$ = speed of sue = 34 m/s

$v_{v}$ = speed of van = 5.20 m/s

$v_{sv}$ = speed of sue relative to van = $v_{s} - v_{v}$ = 34 - 5.20 = 28.8 m/s

$d_{s}$ = stopping distance after brakes are applied

$D$ = distance between sue and van = 160 m

$v_{f}$ = final speed of sue = 0 m/s

$a$ = acceleration = - 1.80 m/s²

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a d_{s}$

$0^{2} = 28.8^{2} + 2 (1.80) d_{s}$

$d_{s} = 230.4$ m

Since $d_{s} < D$

hence there will be collision

7 0

3 years ago

Energy ___________.

AVprozaik [17]

Answer:

BRAINLIEST PLSSSSSSSSSSSSSSSSSSSSSSSS

Thermodynamics

Explanation:

Thermodynamics is the study of energy. First Law of Thermodynamics: Energy can be changed from one form to another, but it cannot be created or destroyed. The total amount of energy and matter in the Universe remains constant, merely changing from one form to another.

7 0

3 years ago

Read 2 more answers

A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done to compress the spring initially= $\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J$

b.

By conservation law of energy

Initial energy of spring=Kinetic energy of object

$37.8=\frac{1}{2}(3)v^2$

$v^2=\frac{37.8\times 2}{3}$

$v=\sqrt{\frac{37.8\times 2}{3}}$

v=5.02 m/s

c.Work done by friction on the incline, $w_{friction}=P.E-spring \;energy$

$W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J$

8 0

3 years ago

A vector has an x-component

NeTakaya

Answer:

34.45m

Explanation:

Magnitude of a vector is equal to the square root of sum of squares of x & y vectors.

Magnitude = $\sqrt({x^2} + {y^2})$

= $\sqrt({19.5^2} + {28.4^2})$

=34.45m

5 0

3 years ago