CO + 7H2 - C3Hg + 3H20

Convert 3.8 kg to Lbs

JulijaS [17]

I believe it’s 8.3 to be precise

8 0

2 years ago

Read 2 more answers

Region that contains the majority of molecules in the atmosphere

Anna71 [15]

Answer:

Trophosphere

Explanation:

The troposphere is the atmospheric layer closest to the planet and is characterized because contains the largest percentage of the mass of the total atmosphere.

On special characteristic is that in this layer the temperature and water vapor content decrease a lot respect to the altitude. Also on this layer the Water vapor is important in order to regulate the air temperature since on this zone we have absorption of the solar energy.

The troposphere contains almost all the water vapor in the atmosphere. And specially on the tropics we have an accumulation of the water vapour.

All weather phenomena occur within the troposphere. Tropos means "change" and Troposphere means "region of mixing".

Above this layer, we have the tropopause, ranges in height from 5 miles near the poles up to 11 miles above the equator. And the height depends of the seasons, with an special characteristic: the is highest height occurs in the summer and lowest height occurs in the winter.

The troposphere contains almost 75% of the mass of the entire atmosphere. The air on this layer is composed by 78% nitrogen, 21% oxygen and 1% is made of argon, water vapor, and carbon dioxide.

So for this reason this is the Region that contains the majority of molecules in the atmosphere.

4 0

3 years ago

Consider a semicircular ring of radius R. Its linear mass density varies as lambda =lambda not sin theta. Locate its centre of m

bearhunter [10]

Answer:

(0, πR/4)

Explanation:

The linear mass density (mass per length) is λ = λ₀ sin θ.

A short segment of arc length is ds = R dθ.

The mass of this short length is:

dm = λ ds

dm = (λ₀ sin θ) (R dθ)

dm = R λ₀ sin θ dθ

The x coordinate of the center of mass is:

X = ∫ x dm / ∫ dm

X = ∫₀ᵖ (R cos θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

X = R ∫₀ᵖ sin θ cos θ dθ / ∫₀ᵖ sin θ dθ

X = R ∫₀ᵖ ½ sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R ∫₀ᵖ 2 sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R (-cos 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

X = ¼R (-cos 2π − (-cos 0)) / (-cos π − (-cos 0))

X = ¼R (-1 + 1) / (1 + 1)

X = 0

The y coordinate of the center of mass is:

Y = ∫ y dm / ∫ dm

Y = ∫₀ᵖ (R sin θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

Y = R ∫₀ᵖ sin² θ dθ / ∫₀ᵖ sin θ dθ

Y = R ∫₀ᵖ ½ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R ∫₀ᵖ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R (θ − ½ sin 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

Y = ½R [(π − ½ sin 2π) − (0 − ½ sin 0)] / (-cos π − (-cos 0))

Y = ½R (π − 0) / (1 + 1)

Y = ¼πR

4 0

3 years ago

180 cm3 of hot tea at 97 °C are poured into a very thin paper cup with 20 g of crushed ice at 0 °C. Calculate the final temperat

Zolol [24]

Answer : The final temperature of the mixture is $91.9^oC$

Explanation :

First we have to calculate the mass of water.

Mass = Density × Volume

Density of water = 1.00 g/mL

Mass = 1.00 g/mL × 180 cm³ = 180 g

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of hot water (liquid) = $4.18J/g^oC$

$c_2$ = specific heat of ice (solid)= $2.10J/g^oC$

$m_1$ = mass of hot water = 180 g

$m_2$ = mass of ice = 20 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of hot water = $97^oC$

$T_2$ = initial temperature of ice = $0^oC$

Now put all the given values in the above formula, we get

$(180g)\times (4.18J/g^oC)\times (T_f-97)^oC=-(20g)\times 2.10J/g^oC\times (T_f-0)^oC$

$T_f=91.9^oC$

Therefore, the final temperature of the mixture is $91.9^oC$

8 0

2 years ago

A girl with a mass of 27 kg is playing on a swing. There are three main forces

N76 [4]

The tension in the swing's chain at the bottom of the swing is 178.35 N.

The given parameters:

Mass of the girl, m = 27 kg
Speed of the girl, v = 3 m/s
Radius of the circle, r = 4 m

The tension in the swing's chain at the bottom of the swing is calculated as follows;

$T = mg + ma_c\\\\ T= mg + \frac{mv^2}{r} \\\\T = (12 \times 9.8) + (\frac{27 \times 3^2}{4} )\\\\T = 117.6 \ N \ + \ 60.75 \ N\\\\T = 178.35 \ N$

Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.

Learn more about tension in vertical circle here: brainly.com/question/19904705

3 0

2 years ago