Fruits and vegetables are in the produce aisle because they are plants, and plants are producers.
Producers are organisms that create energy on their own through various processes depending on the organism.

6 0

3 years ago

Pls help me on this it’s past due ty

mihalych1998 [28]

Answer:

sorry

Explanation:

6 0

2 years ago

A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235

trapecia [35]

Answer : The value of $\Delta E$ of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

$\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}$

Molar mass of helium = 4 g/mole

$\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole$

Now we have to calculate the heat.

Formula used :

$q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)$

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

$c_p$ = specific heat of helium = 20.8 J/mol.K

$T_1$ = initial temperature = 310.0 K

$T_2$ = final temperature = 235.0 K

Now put all the given values in the above formula, we get:

$q=nc_p(T_2-T_1)$

$q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K$

$q=-19.5J$

Now we have top calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 1.21 L

$V_2$ = final volume = 0.99 L

Now put all the given values in the above formula, we get:

$w=-p(V_2-V_1)$

$w=-(1atm)\times (0.99-1.21)L$

$w=0.22L.artm=0.22\times 101.3J=22.29J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the value of $\Delta E$ of the gas.

$\Delta E=q+w$

$\Delta E=(-19.5J)+22.29J$

$\Delta E=2.79J$

Therefore, the value of $\Delta E$ of the gas is 2.79 Joules.

3 0

2 years ago

Suppose that on a dry, sunny day when the air temperature is near 37 ∘C,37 ∘C, a certain swimming pool would increase in tempera

ioda

Answer:

The fraction of water body necessary to keep the temperature constant is 0,0051.

Explanation:

Heat:

$Q=m*Ce*ΔT$

Q= heat (unknown)

m= mass (unknown)

Ce= especific heat (1 cal/g*°C)

ΔT= variation of temperature (2.75 °C)

Latent heat:

$ΔE=∝mΔHvap$

ΔE= latent heat

m= mass (unknown)

∝= mass fraction (unknown)

ΔHvap= enthalpy of vaporization (539.4 cal/g)

Since Q and E are equal, we can match both equations:

$m*Ce*ΔT=∝*m*ΔHvap$

Mass fraction is:

$∝=\frac{Ce*ΔT}{ΔHvap}$

$∝=\frac{(1 cal/g*°C)*2.75°C}{539.4 cal/g}$

∝=0,0051

7 0

2 years ago