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r-ruslan [8.4K]

2 years ago

3

Rina’s mother asked her to help move some items from the hall closet to the garage. One item she wanted to move was a large box

of old school books. Rina tried to lift the box, but could not pick it up. What force was Rina NOT able to overcome?

2 answers:

RoseWind [281]2 years ago

6 0

Answer:

is it gravity?? I think it is

Marat540 [252]2 years ago

4 0

Answer:

I would say gravitational force

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When atoms lose or gain electrons in chemical reactions they form?

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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

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2 years ago

What is the element with the nlx notation of 5d2?

lesantik [10]

The element with the nlx notation of 5d2 in the periodic table is hafnium with a symbol of hf. It is a chemical element and has an atomic number 72. H<span>afnium chemically resembles zirconium and is found in many zirconium minerals but it is a complete contrast of zirconium which is also chemically similar to hafnium.</span>

7 0

2 years ago

This is the chemical formula for methyl acetate: CH32CO2. Calculate the mass percent of hydrogen in methyl acetate. Round your a

WITCHER [35]

<u>Answer:</u> The mass percent of hydrogen in methyl acetate is 8 %

<u>Explanation:</u>

The given chemical formula of methyl acetate is $CH_3COOCH_3$

To calculate the mass percentage of hydrogen in methyl acetate, we use the equation:

$\text{Mass percent of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of methyl acetate}}\times 100$

Mass of hydrogen = (6 × 1) = 6 g

Mass of methyl acetate = [(3 × 12) + (6 × 1) + (2 × 16)] = 74 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen}=\frac{6g}{74g}\times 100=8.10\%=8\%$

Hence, the mass percent of hydrogen in methyl acetate is 8 %

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2 years ago

At what power of ten are we able to view the entire Earth in space?

umka21 [38]

Answer:

If I remember correctly it should be 10^7 meters.

Explanation:

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2 years ago