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9966 [12]
2 years ago
15

Ben Tooclose is being chased through the woods by a bull moose which he attempting to photograph.the enormous mass of the bull m

oose is extremely intimidating.yet, if ben makes a zigzag pattern through the woods, he will be able to use the large mass of the moose to his own advantage.explain this in terms of inertia and newton's first law of motion.
Chemistry
1 answer:
Elodia [21]2 years ago
4 0
Newton's first law of motion is the law of inertia. There are basically "two parts" to the law of inertia. The first part says that an object at rest tends to remain at rest unless it is acted on by an outside force. Think of gravity.
-Hope that helps!! :)
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Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll
makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature T_1 = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}

where; B = - 152.5 \ cm^3 /mol   C = -5800 cm^6/mol^2

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

4.138*10^{-4}  \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

Multiplying through with V² ; we have

4.138*10^4  \ V ^3 = V^2 - 152.5 V - 5800 = 0

4.138*10^4  \ V ^3 - V^2 + 152.5 V + 5800 = 0

V = 2250.06  cm³ mol⁻¹

Z = \frac{PV}{RT}

Z = \frac{1800*2250.06}{8.314*10^3*523.15}

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

T_c = 647.1 \ K \\ \\ P_c = 22055 \  kPa  \\ \\ \omega = 0.345

T__{\gamma}} = \frac{T}{T_c}

T__{\gamma}} = \frac{523.15}{647.1}

T__{\gamma}} = 0.808

P__{\gamma}} = \frac{P}{P_c}

P__{\gamma}} = \frac{1800}{22055}

P__{\gamma}} = 0.0816

B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}

B_o = 0.083 - \frac{0.422}{0.808^{1.6}}

B_o = 0.51

B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}

B_1 = -0.282

The compressibility is calculated as:

Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}

Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}

Z = 0.9386

V= \frac{ZRT}{P}

V= \frac{0.9386*8.314*10^3*523.15}{1800}

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At T_1 = 523.15 \  K \ and  \ P = 1800 \ k Pa

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = \frac{PV}{RT}

Z = \frac{1800*10^3 *0.1249}{729.77*523.15}

Z = 0.588

3 0
2 years ago
A 6.0M solution HCl is diluted to 1.0M How many milliliters of the 6.0M solution would be used to prepare 100.o mL of the dilute
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Wats r is the most soluble object ever. It is dense
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. Divide 94.20 g by 3.167 22 mL.
jeka94
94.20 g/3.16722 mL = 29.74 g/mL

The ratio of mass to volume is equal to the substance's density. Thus, 29.74 g/mL is the density of whatever substance it may be. Density does not change for incompressible matter like solid and some liquids. Although, it may be temperature dependent.
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What is the fraction of the hydrogen atom's mass (11h) that is in the nucleus? the mass of proton is 1.007276 u, and the mass of
ololo11 [35]
Hello there!

To determine the fraction of the hydrogen atom's mass that is in the nucleus, we have to keep in mind that a Hydrogen atom has 1 proton and 1 electron. Protons are in the nucleus while electrons are in electron shells surrounding the nucleus. The mass of the nucleus will be equal to the mass of 1 proton and we can express the fraction as follows:

Mass Fraction= \frac{mass 1 Proton}{mass H atom} = \frac{1,007276 u}{1,007825}=0,9995

So, the fraction of the hydrogen atom's mass that is in the nucleus is 0,9995. That means that almost all the mass of this atom is at the nucleus.

Have a nice day!
3 0
2 years ago
Metals are considered this if they can be made into sheets
Korvikt [17]

They are considered malleable. They can be made into sheets

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