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3 years ago

What is/are the products in the chemical equation?

Physics

2 answers:

Gennadij [26K]3 years ago

8 0

$\huge{ \mathfrak{ \underline{ Answer }}}࿐$

The balanced Chemical equation for above reaction is :

$\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$

The product formed here is $NH_3$ (Ammonia)

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$\#TeeNForeveR$

Alla [95]3 years ago

5 0

Answer:NH3

Explanation: product s are the result of the equation

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Canada’s northern territories experience "midnight sun," meaning what?.

Leona [35]

Answer:

Explanation: The term "midnight sun" refers to the consecutive 24-hour periods of sunlight experienced in the north of the Arctic Circle and south of the Antarctic Circle. Other phenomena are sometimes referred to as "midnight sun", but they are caused by time zones and the observance of daylight saving time.

6 0

3 years ago

A bungee jumper of mass 75kg is attached to a bungee cord of length L=35m. She walks oﬀ a platform (with no initial speed), reac

attashe74 [19]

Answer:

1. 77.31 N/m

2. 26.2 m/s

3. increase

Explanation:

1. According to the law of energy conservation, when she jumps from the bridge to the point of maximum stretch, her potential energy would be converted to elastics energy. Her kinetic energy at both of those points are 0 as speed at those points are 0.

Let g = 9.8 m/s2. And the point where the bungee ropes are stretched to maximum be ground 0 for potential energy. We have the following energy conservation equation

$E_P = E_E$

$mgh = kx^2/2$

where m = 75 kg is the mass of the jumper, h = 72 m is the vertical height from the jumping point to the lowest point, k (N/m) is the spring constant and x = 72 - 35 = 37 m is the length that the cord is stretched

$75*9.8*72 = 37^2k/2$

$k = (75*9.8*72*2)/37^2 = 77.31 N/m$

2. At 35 m below the platform, the cord isn't stretched, so there isn't any elastics energy, only potential energy converted to kinetics energy. This time let's use the 35m point as ground 0 for potential energy

$mv^2/2 = mgH$

where H = 35m this time due to the height difference between the jumping point and the point 35m below the platform

$v^2/2 = gH$

$v = \sqrt{2gH} = \sqrt{2*9.8*35} = 26.2 m/s$

3. If she jumps from her platform with a velocity, then her starting kinetic energy is no longer 0. The energy conservation equation would then be

$E_P + E_k = E_E$

So the elastics energy would increase, which would lengthen the maximum displacement of the cord

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4 years ago

The velocity of a body of mass 20kg decreases from 20m/s to 5m/s in a distance of 100 m force on body is ?

MrRissso [65]

Answer:

hope it helps brainliest pls. and the answer is the one that is circled in red color.

Explanation:

8 0

3 years ago

Calculate the pressure exerted by 11.1 moles of neon gas in a volume of 5.45 L at 25°C using (a) the ideal gas equation and (b)

Ilia_Sergeevich [38]

Answer:

49.82414 atm

50.74675 atm

Explanation:

P = Pressure

V = Volume = 5.45 L

R = Gas constant = 0.08205 L atm/mol K

T = Temperature = 25°C

a = 0.211 atm L²/mol²

b = 0.0171 atm L²/mol²

From ideal gas law we have

$PV=nRT\\\Rightarrow P=\dfrac{nRT}{V}\\\Rightarrow P=\dfrac{11.1\times 0.08205(273.15+25)}{5.45}\\\Rightarrow P=49.82414\ atm$

The pressure is 49.82414 atm

From Van der Waals equation we have

$\left(P+\frac{an^2}{V^2}\right)\left(v-nb\right)=nRT\\\Rightarrow P=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\\\Rightarrow P=\dfrac{11.1\times 0.08205\times (273.15+25)}{5.45-(11.1\times 0.0171)}-\dfrac{0.211\times 11.1^2}{5.45^2}\\\Rightarrow P=50.74675\ atm$