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melamori03 [73]
3 years ago
8

Help plz:)))I’ll mark u Brainliest

Chemistry
1 answer:
finlep [7]3 years ago
7 0

Answer:

11690J

Explanation:

Use formula Q = mL

where:

Q = Energy (J)

m = Mass (g)

L = Latent heat of fusion for water (J/g) = 344J/g

Therefore:

Q = 35.0 x 344

Q = 11690J

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Is the sweetened iced tea an electrolyte mixture?
Arada [10]

Answer:

Yes it is

Explanation:

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How many molecules are in the quantities below.? 2.0 moles
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Mole=number of molecules/6.02x10²³
mole=2
number of molecules= 2x6.02x10²³
number of molecules=12.04x10²³
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What is the amount of heat released when 25g of water cools 12.5 degrees C ?
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Water can't cool at a single temperature. It must start at a higher temperature, and drop to a lower temperature in order to cool. Unless we know the other temperature, there is no way to calculate the amount of thermal energy released.
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Suppose we have two rock samples, A and B. If rock A was subject to both physical and chemical weathering while rock B was subje
hammer [34]

Answer:

Rock A because after physical weather and chemical weathering it more likely for Rock A to experience more chemical weathering.

Explanation: Weathering: This is a geological term used to describe the various processes and Activities involved in the breaking down of rocks either through physical,mechanical,chemical etc actions into smaller particles.

ROCK A WILL HAVE MORE CHEMICAL WEATHERING BECAUSE THE PHYSICAL WEATHERING MUST HAVE BROKEN DOWN THE PARTICLES FOR EASY WATER AND OTHER SUBSTANCE NEEDED FOR EASIER CHEMICAL REACTION OR WEATHERING.

4 0
3 years ago
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m
kogti [31]

Answer:

The molar mass of the unknown gas is \mathbf{ 51.865 \  g/mol}

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

R \  \alpha  \ \dfrac{1}{\sqrt{d}}

R = \dfrac{k}{d}  where K = constant

If we compare the rate o diffusion of two gases;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}

Since the density of a gas d is proportional to its relative molecular mass M. Then;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}

Rate is the reciprocal of time ; i.e

R = \dfrac{1}{t}

Thus; replacing the value of R into the above previous equation;we have:

\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}

We can equally say:

{\dfrac{t_2}{t_1}}=  {\sqrt{\dfrac{M_2}{M_1}}

{\dfrac{6.34}{4.98}}=  {\sqrt{\dfrac{M_2}{32}}

M_2 = 32 \times ( \dfrac{6.34}{4.98})^2

M_2 = 32 \times ( 1.273092369)^2

M_2 = 32 \times 1.62076418

\mathbf{M_2 = 51.865 \  g/mol}

7 0
3 years ago
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