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3 years ago

Please help!!!

Chemistry

1 answer:

oee [108]3 years ago

3 0

Answer:

Please help!!!

When a piston with a volume of 35 mL is heated from 298 K to 596 Kit

expands. Assuming the pressure on the piston remains constant, determine

the new volume of the cylinder. (round to the nearest whole number)

Explanation:

n0? = °^° hhaa jk

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SVETLANKA909090 [29]

11 is the answer hope that helps

4 0

3 years ago

A car travelling at a uniform speed of 100 kmh spends 15minutes moving from point A to point B along its route. the distance bet

gizmo_the_mogwai [7]

Answer:

25002m

Explanation:

V = 100 km/h

Converting to m/s, we have

V = (100 x 1000)/3600 = 27.78m/s

t = 15mins = 15 x 60 = 900secs

Displacement =?

Velocity = Displacement /Time

Displacement = Velocity x time

Displacement = 27.78 x 900

Displacement = 25002m

Since displacement and distance are measured in same unit, the distance between A and B is 25002m

7 0

3 years ago

You are burning wood to heat water for your industrial process. What is the mass of wood required to raise the temperature of 10

natta225 [31]

Answer:

18,8kg of wood

Explanation:

The energy you need to to raise the temperature of 1000 kg of water from 25.0 to 100.0 °C is:

q = C×m×ΔT

Where: q is heat, C is specific heat of water (4,184J/g°C), m is mass in grams (1000x10³g), and ΔT is 100,0°C - 25,0°C = 75,0°C

Replacing:

q = 4,184J/g°C×1000x10³g×75,0°C

q = 3,14x10⁸ J of heat are required

Now, if the heating value of dry wood is 16,72 MJ/kg = 16,72x10⁶ J/kg, mass of wood required is:

3,14x10⁸J × (1kg / 16,72x10⁶ J) = 18,8 kg of wood are required

I hope it helps!

5 0

3 years ago

Which part of a feedback mechanism causes change to make up for the departure from the set point? A. sensor B. effector C. respo

klemol [59]

B.) Effector
I believe so lol

6 0

3 years ago

calculate the freezing point of 3.46 gram of a compound X in 160 gram of benzene when a separate sample of X was vaporized it's

Temka [501]

Answer: The freezing point of 3.46 gram of a compound X in 160 gram of benzene is $4.38^0C$

Explanation:

The relation of density and molar mass is:

$d=\frac{PM}{RT}$

where

d = density = 3.27 g/ L

P = pressure of the gas = 773 torr = 1.02 atm (760 torr = 1atm)

M = molar mass of the gas = ?

T = temperature of the gas = $116^0C=(116+273)K=389K$

R = gas constant = $0.0821Latm/Kmol$

$M=\frac{dRT}{P}=\frac{3.27g/L\times 0.0821Latm/Kmol\times 389K}{1.02atm}=102.3g/mol$

The relation of depression in freezing point with molality:

$\Delta T_f=k_f\times m$

$\Delta T_f$ = depression in freezing point = $T_f^0-T_f$ = $5.45-T_f$

$k_f$ = freezing point constant = 5.1

m = molality = $\frac{\text {moles of X}}{\text {weight of solvent in kg}}=\frac{3.46\times 1000}{102.3\times 160}=0.21$

$5.45-T_f=5.1\times 0.21$

$T_f=4.38^0C$

Thus the freezing point of 3.46 gram of a compound X in 160 gram of benzene is $4.38^0C$

5 0

3 years ago