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3 years ago

A salt is dissolved in water which has a freezing point of 0 degrees celsius. the freezing point of the solution would be

2 answers:

7 0

A salt is dissolved in water which has a freezing point of 0 degrees celsius. the freezing point of the solution would be dependent on the concentration of the salt in the solution. It is explained by the colligative properties. These are properties that depend upon the concentration of solute molecules or ions, but not upon the identity of the solute. Hope this answers the question.

MrMuchimi3 years ago

7 0

Hello there.

A salt is dissolved in water which has a freezing point of 0 degrees celsius. the freezing point of the solution would be

solution would be dependent on the concentration of the salt in the solution.

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Given a force of 50 N and an acceleration of 5 m/s/s, what is the object’s mass?

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3 years ago

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Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of eac

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Answer:

t = (ti)ln(Ai/At)/ln(2)

t = 14ln(16)/ln(2)

Solving for t

t = 14×4 = 56 seconds

Explanation:

Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t

At = Ai/2^n ..... 1

Where n is the number of half-life that have passed.

n = t/half-life

Half life = 14

n = t/14

At = Ai/2^(t/14)

From equation 1.

2^n = Ai/At

Taking the natural logarithm of both sides;

nln(2) = ln(Ai/At)

n = ln(Ai/At)/ln(2)

Since n = t/14

t/14 = ln(Ai/At)/ln(2)

t = 14ln(Ai/At)/ln(2)

Ai = 800

At = 50

t = 14ln(800/50)/ln(2)

t = 14ln(16)/ln(2)

Solving for t

t = 14×4 = 56 seconds

Let half life = ti

t = (ti)ln(Ai/At)/ln(2)

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4 years ago

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2 years ago

Which tool can be used to measure the volume of a liquid to one decimal place?

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4 years ago

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A 8.0-cm-diameter horizontal pipe gradually narrows to 5.0 cm . When water flows through this pipe at a certain rate, the gauge

adelina 88 [10]

Answer:

A 8.0 cm diameter horizontal pipe gradually narrows to 5.0 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

The flow rate is 3.1175×10⁻³ m³/s

Explanation:

To solve the question we rely on Bernoulli's principle as follows $P_{1} +\frac{1}{2}\rho v^{2} _{1} + \rho gz_{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2} + \rho gz_{2}$

thus where the pipe is horizontal we have

z₁ = z₂ hence the above equattion becomes

$P_{1} +\frac{1}{2}\rho v^{2} _{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2}$

since the flow rate is constant then

Q = v₁A₁ = v₂A₂

Where is the area of the two sections given by A₁ = π·D₁²÷4 and

A₂ = π·D₂²÷4

Thereffore A₁ = π·0.08²÷4 = 5.02×10⁻³ m²

and A₂ = π·0.05²÷4 = 1.96×10⁻³ m²

v₁ = v₂A₂/A₁ =0.391×v₂

The given pressures are P₁ = 31.0 kPa and P₂ = 24.0 pKa and

ρ = 1000 kg/m³

Plugging the values into the above equation we get

31.0 kPa +0.5× 1000 kg/m³× (0.391×v₂)² = 24.0 pKa +0.5×1000 kg/m³×v₂²

= 31000+76.3·v₂² =24000+500·v₂²

or 423.706·v₂² = 7000

v₂² = 7000/423.706 = 16.52 or v₂ = 4.065 m/s and v₁ 0.391×4.065 = 1.59 m/s

The flow rate = v₂A₂ = 1.59×1.96×10⁻³ = 3.1175×10⁻³ m³/s

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