answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
nydimaria [60]
2 years ago
11

A particular sheet of paper measures 8.5 × 6.5 inches. What is the surface area of one side of the paper in cm2? (2.54 cm = 1 in

exactly)
Chemistry
1 answer:
Zanzabum2 years ago
4 0
<span>In order to calculate the surface are of the sheet of paper in square centimeters you must first convert the dimensions of the paper to centimeters. You can do this by multiplying 8.5 * 2.54 and 6.5 * 2.54. The paper measures 21.59cm by 16.51cm. To find the surface area you multiply the dimensions of the paper to equal 356.4509cm2</span>
You might be interested in
In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)
Maru [420]

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.6 L  

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm (From correct source)

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}

{V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L

{V_2}=0.27\ L

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

7 0
2 years ago
A sample of an unknown substance has a mass of .158 kg if 2,510 J of heat is required to heat the substance from 32°C to 61°C wh
ziro4ka [17]

Answer: 0.548J/g°C

Explanation:

Q = s × m × DeltaT

Q = Heat (J)

S = Specific Heat Capacity

M = mass (g)

DeltaT = Change in temperature (°C)

0.158Kg x 1000 = 158g

2.510J = s x 158g x (61°C-32°C)

2.510J/(158g x 29°C) = s

S = 0.54779.... J/g°C

S = 0.548 J/g°C

7 0
2 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
The reading on a speedometer represents the average velocity of the car. True False
Sedbober [7]
Answer: False. Hope this helps!!
7 0
2 years ago
Some household cleaners come in concentrations stronger than necessary for basic cleaning jobs. Jeremy followed the instructions
VashaNatasha [74]

Answer:

no he just repeated the steps and made more of the same cleaner my guy

Explanation:

5 0
2 years ago
Other questions:
  • Determine the mass of oxygen in a 7.20 g sample of Al2(SO4)3.
    9·2 answers
  • A 25.0 ml sample of an unknown hbr solution is titrated with 0.100 m naoh. the equivalence point is reached upon the addition of
    7·1 answer
  • Compound b (c9h9no4) is a carboxylic acid that has a nitro group attached to an aromatic ring. draw this compound based on its 1
    9·1 answer
  • 40pionts
    9·1 answer
  • A beaker contains a dilute sodium chloride solution at 1 atmosphere. What happens to the number of solute particles in the solut
    6·2 answers
  • A 37.2-g sample of lead (Pb) pellets at 20°C is mixed with a 62.7-g sample of lead pellets at the same temperature. What are th
    10·1 answer
  • (a) The mass density of a gaseous compound was found to be 1.23 kg m^−3 at 330 K and 20 kPa. What is the molar mass of the compo
    8·1 answer
  • A sample of krypton has a volume of 6.00 L, and the pressure is 0.960 atm. If the final temperature is 55.0°C, the final volume
    11·1 answer
  • A glass cup of orange juice is found to have pOH of 11.40. Calculate the concentration of the hydrogen ions in the juice.​
    15·1 answer
  • A student in a chemistry laboratory has access to two acid solutions. The first one is 20% acid and the second solution is 45% a
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!