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2 years ago

5

A system is a group of related parts with specific roles that work together to achieve an observed result. Seeds, soil, a plow,

and a large field are parts of what type of system?
A.
an agricultural system
B.
an organ system
C.
a legal system
D.
an urban system

2 answers:

ser-zykov [4K]2 years ago

6 0

Answer - A. an agricultural system

Explanation - Anything that has two or more parts working together is a type of system. Some systems, such as toothbrushes, are very simple because they only have a few parts. Other systems, such as farms, can be very complex because they have hundreds of different parts. The components of a system play different roles in the system.

Plus study island:D

VMariaS [17]2 years ago

3 0

The answer is A, i'm also on study island on this test. It has a star next to it. So yep we're right!

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What is the number of moles of <br> 0.0960g of H2SO4

Vilka [71]

<span>Answer</span>=.000978802802moles H2SO4

How I Got My Answer

<span>Molar mass
</span>H2SO4= 98.079g/mol

What I have
.0960g H2SO4

Equation
.0960g*1mol/98.079g= .000978802802mol H2SO4

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2 years ago

For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugati

AveGali [126]

Answer:

a. withdraws electrons inductively

b. donates electrons by hyperconjugation

c. donates electrons by resonance

d. withdraws electrons inductively

Explanation:

a. The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:

F > Cl > Br>I

b. The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons. Because of the delocalization, hyperconjugation is the result effect.

c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.

d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.

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2 years ago

Which nuclear emission moving through an electric field would be deflected toward the positive electrode?

Tom [10]

As a rule, opposite charges attract. So,you would expect that the particle that would deflect towards the positive electrode has a negative charge. Now, an alpha particle is basically a neutral Helium atom. Beta particle is an electron. Gamma particle has no charge (neutral). <em>Thus, the answer is B.</em>

6 0

3 years ago

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

What is the percent CdSO4 by mass in a 1.00 m aqueous CdSO4 solution?

Svet_ta [14]

I don't know but I'm wasting 5 seconds of your time you can't take back sorry

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2 years ago