Are the “strings” of celery xylem or pholem tissue?

HELP ME WITH ONE OR BOTH OF THESE QUEATIONS PLEASEEEE

marta [7]

Answer:

2. 181.25 K.

3. 0.04 atm.

Explanation:

2. Determination of the temperature.

Number of mole (n) = 2.1 moles

Pressure (P) = 1.25 atm

Volume (V) = 25 L

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =?

The temperature can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

1.25 × 25 = 2.1 × 0.0821 × T

31.25 = 0.17241 × T

Divide both side by 0.17241

T = 31.25 / 0.17241

T = 181.25 K

Thus, the temperature is 181.25 K.

3. Determination of the pressure.

Number of mole (n) = 10 moles

Volume (V) = 5000 L

Temperature (T) = –10 °C = –10 °C + 273 = 263 K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =?

The pressure can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

P × 5000 = 10 × 0.0821 × 263

P × 5000 = 215.923

Divide both side by 5000

P = 215.923 / 5000

P = 0.04 atm

Thus, the pressure is 0.04 atm

6 0

2 years ago

A gas occupies 3.5L at 2.5 kPa pressure. What is the volume at 100 mmHg at the same temperature? Be sure to

nasty-shy [4]

The volume at 100 mmHg : 0.656 L

<h3>Further explanation</h3>

Boyle's Law

<em>At a constant temperature, the gas volume is inversely proportional to the pressure applied </em>

$\rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}$

V₁=3.5 L

P₁=2.5 kPa=18,7515 mmHg

P₂=100 mmHg

$\tt \dfrac{18.7515}{100}=\dfrac{V_2}{3.5}\\\\V_2=\dfrac{18.7515\times 3.5}{100}=0.656~L$

4 0

2 years ago

A fluid occupying has a mass of 4mg. Calculate its density and specific volume in SI, EE, and BG units.

kondaur [170]

The question is incomplete, complete question is:

A fluid occupying $3.2 m^3$ of volume has a mass of 4 Mg. Calculate its density and specific volume in SI, EE, and BG units.

Explanation:

1) Mass of liquid = m = 4 Mg = 4 × 1,000 kg = 4,000 kg

(1 Mg = 1000 kg)

Volume of the fluid = V = $3.2 m^3$

Density of the fluid = D

$D=\frac{m}{V}=\frac{4,000 kg}{3.2 m^3}=1,250 kg/m^3$

Specific volume is the reciprocal of the density :

$V_{specific}=\frac{1}{Density}$

Specific volume of the fluid = $S_v$

$S_v=\frac{1}{D}=\frac{1}{1,250 kg/m^3}=0.0008 m^3/kg$

2)

Density of the fluid in English Engineering units = D $(lb/ft^3)$

1 kg = 2.20462 lb

1 m = 3.280 ft

$D=\frac[1,250\times 2.20462 lb}{(3.280 ft)^3=78.95 lb/ft^3$

Specific volume of the fluid :

$=\frac{1}{78.95 lb/ft^3}=0.0127 ft^3/lb$

3)

Density of the fluid in British Gravitational System units = D $(slug/ft^3)$

1 kg = 0.06852 slug

1 m = 3.280 ft

$D=\frac[1,250\times 0.0685218 slug}{(3.280 ft)^3=2.43 slug/ft^3$

Specific volume of the fluid :

$=\frac{1}{2.43 slug/ft^3}=0.412 ft^3/slug$

7 0

2 years ago

William adds two values , following the rules for using significant figures in computations. He should write the sum of these tw

yuradex [85]

Answer:

when it comes to adding or subtracting numbers, his final answer should have the same number of decimal places as the least precise value.

For example if you add 2 numbers; 10.443 + 3.5 , 10.443 has 3 decimal places and 3.5 has only one decimal place.

Therefore 3.5 is the less precise value.

So when adding these 2 values the final answer should have only one decimal place.

after adding we get 13.943 but it can have upto one decimal place. then the second decimal place is less than 5 so the answer should be rounded off to 13.9.

the answer is the same number of decimal places as the least precise value

Explanation:

I think this is the answer I'm not sure

7 0

2 years ago

Pls helps and no files since my school have blocked things like that on my computer

Lubov Fominskaja [6]

I did this the answer is B

3 0

2 years ago

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