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2 years ago

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If an object has more positive charges than negative charges, the object has an overall?

1 answer:

djyliett [7]2 years ago

7 0

The object has an overall positive charge.

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Use the drop-down menus to select the correct name for each of the organic compounds.

Molodets [167]

Answer:

Here's what I get

Explanation:

CH₃CH₂CH₂CH₂CH₂CH₃ — hexane

CH₂=CHCH₂CH₂CH₂CH₃ — hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted

CH₃C≡CCH₃ — but-2-yne (PIN); 2-butyne is accepted

CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane

CH₃CH₂CHCICH₂CH₃ — 3-chloropentane

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3 years ago

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P l e a s e help me omg I'm so confused-I'll give 40 points, I only have 45-

stiks02 [169]

M is not as thick as N, It only takes 100°C to boil water whereas it takes 300°C to boil soybean oil

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2 years ago

Which of the following is an example of a compound? A. helium B. water C. hydrogen B. oxygen

liq [111]

the compound is B.

Water is a compound because it is made up of water molecules.

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2 years ago

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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

Chuẩn Độ 15 ml dung dịch CH3COOH 0,2 m bằng dung dịch NaOH 0,2 m

mina [271]

Answer:

<h3>I don't know what is the answer of your question sorry never mind..</h3>

Explanation:

<h3>And please marks me as brainliest... </h3>

3 0

2 years ago