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miv72 [106K]
3 years ago
8

What is 1 Oz equal to in grams?

Chemistry
2 answers:
lakkis [162]3 years ago
4 0
1 Oz is 28.3495 grams

hope this helps!
pentagon [3]3 years ago
3 0

Conversion factor for ounces and grams.

<em>1 ounce is equal to approximately 28.4 grams. </em>

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5000kg of ammonium nitrate per square kilometer of cornfield per year. how much nitric acid would be needed to make the fertiliz
Serggg [28]

NH_3+HNO_3-> NH_4NO_3

1 mole of nitric acid produce 1 mole of ammonium nitrate.

moles in 5000 kg of ammonium nitrate :

n=\dfrac{5000\times 1000}{80}\\\\n=62500\ moles ( molecular mass of ammonium nitrate is 80 gm/mol )

So, number of moles of nitric acid required are also 62500 moles.

Mass of 62500 moles of nitric acid :

mass = 62500\times 63 \\\\mass = 3937500\ gram\\\\mass = 393.75\ kg

Hence, this is the required solution.

5 0
2 years ago
How can radioactivity be used to help find the age of an object ?
gogolik [260]

Hope this answers your question Mariaduong159

Radiometric Dating. It's used to find the dates of ricks and other objects based on what the known decay rate of radioactive isotopes. Different forms of this method can also estimate the age of natural and man-made materials.

8 0
3 years ago
Soil particles tend to be __________ charged and attract __________ charged ions.
aliina [53]

Answer:it’s abc it’s just science you know

Explanation:

8 0
3 years ago
Why do atoms form blonds​
lakkis [162]

Answer:

Atoms form chemical bonds to make their outer electron shells more stable. ... An ionic bond, where one atom essentially donates an electron to another, forms when one atom becomes stable by losing its outer electrons and the other atoms become stable (usually by filling its valence shell) by gaining the electrons.

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7 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
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