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grandymaker [24]
2 years ago
5

Please answer

Chemistry
1 answer:
AURORKA [14]2 years ago
4 0

Answer:

hi

Explanation:

1 east

2 the flower bud for cotton and for jute the leaves

plz mark me as a BRAINLIST

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What is an individual living entity called?
lions [1.4K]
In the most simplest answer it is organism
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1 year ago
A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num
-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}

<em>B) Radius</em>

\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}

V = \frac{ 4}{3 }\pi r^{3}

r^{3} = \frac{3V }{4 \pi }\

r= \sqrt [3]{ \frac{3V }{4 \pi } }

r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}

3 0
3 years ago
Exactly one mole of an ideal gas is contained in a 2.00 liter container at 1,000 K. What is the pressure exerted by this gas?
Kipish [7]
pV = nRT

p = nRT/V 

p= 1 x 0.08205 x 1000/ 2

p = 41.025 Pa  

Edit: The unit should be atm instead of Pa, as pointed out by a nice human being.
4 0
3 years ago
Read 2 more answers
pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which beha
Fiesta28 [93]

Complete Question

Questions Diagram is attached below

Answer:

*  W=1142.86Joule

*  Q=997.7J

*  H=2140.5J

Explanation:

From the question we are told that:

Temperature T=337K

Pressure P=(60-55)Pa*10^5

VolumeV=(1.6-1.4)m^3*10^{-3}

Generally the equation for gas Constant is mathematically given by

\frac{P_2}{P_1}=\frac{V_1}{V_2}^n

 \frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n

 n=0.65

Therefore

Work-done

 W=\int{pdv}

 W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}

 W=1142.86Joule

Generally the equation for internal energy is mathematically given by

 Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}

 Q=997.7J

Therefore

 H=Q+W

 H=997.7J-11.42.9

 H=2140.5J

5 0
2 years ago
Dissolution of wax in kerosene
yanalaym [24]

Answer:

the first answer is correct don't forget you can use quizzlet app to

8 0
3 years ago
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