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3 years ago

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How much heat energy is required to raise the temperature of 50g of bromine from 25°C to 30°C? [Specific heat capacity of bromin

e = 0.226 J/(g °C]

1 answer:

tia_tia [17]3 years ago

6 0

Answer:

56.5J

Explanation:

<em>To </em><em>find</em><em> </em><em>the </em><em>heat </em><em>energy</em><em> </em><em>required</em><em> </em><em>use </em><em>the </em><em>formula</em><em> </em><em>for </em><em>the </em><em>specific</em><em> </em><em>heat </em><em>capacity</em><em> </em><em>which </em><em>is </em>

<em>c=</em><em>quantity</em><em> of</em><em> </em><em>heat/</em><em>mass×</em><em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>in </em><em>temperature</em>

<em>in </em><em>this </em><em>question</em><em> </em><em>c </em><em>is </em><em>0</em><em>.</em><em>2</em><em>2</em><em>6</em><em>j</em><em>/</em><em>g,</em><em>the </em><em>mass </em><em>is </em><em>5</em><em>0</em><em>g</em><em> </em><em>and </em><em>the </em><em>change </em><em>in </em><em>temperature</em><em> </em><em>is </em><em>3</em><em>0</em><em>-</em><em>2</em><em>5</em><em>=</em><em>5</em>

<em>therefore</em>

<em>0</em><em>.</em><em>2</em><em>2</em><em>6</em><em>=</em><em>Q/</em><em>5</em><em>0</em><em>×</em><em>5</em>

<em>Q=</em><em>0</em><em>.</em><em>2</em><em>2</em><em>6</em><em>×</em><em>2</em><em>5</em><em>0</em>

<em> </em><em> </em><em> </em><em>=</em><em>56.5J</em>

<em>I </em><em>hope </em><em>this</em><em> helps</em>

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Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

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Answer:

Follows are the solution:

Explanation:

A + B = C

Its response decreases over time as well as consumption of a reactants.

r = -kAB

during response A convert into 2x while B convert into x to form 3x of C

let's y = C

y = 3x

Still not converted sum of reaction

for A: 100 - 2x

for B: 50 - x

Shift of x over time

$\frac{dx}{dt} = \frac{-k(100 - 2x)}{(50 - x)}$

Integration of x as regards t

$\frac{1}{[(100 - 2x)(50 - x)]} dx = -k dt\\\\\frac{1}{2[(50 - x)(50 - x)]} dx = -k dt\\\\\ integral\ \frac{1}{2[(50 - x)^2]} dx =\ integral [-k ] \ dt\\\\\frac{-1}{[100-2x]} = -kt + D \\\\$

D is the constant of integration

initial conditions: t = 0, x = 0

$\frac{-1}{[100-2x]} = -kt + D \\\\\frac{ -1}{[100]} = 0 + D\\\\D= \frac{-1}{100}\\\\$

hence we get:

$\frac{-1}{[100-2x]}= -kt -\frac{1}{100}\\\\or \\\\ \frac{1}{(100-2x)} = kt + \frac{1}{100}$

after t = 7 minutes , $C = 10 \ g = 3x$

$3x = 10\\\\x = \frac{10}{3}$

Insert the above value x into $\frac{1}{(100-2x)}$ equation $= kt + \frac{1}{100}$ to get k.

$\to \frac{1}{(100-2\times \frac{10}{3})} = k \times (7) + \frac{1}{100} \\\\ \to \frac{1}{(100- 2 \times 3.33)} = \frac{700k + 1}{100} \\\\ \to \frac{1}{(100-6.66)} = \frac{700k + 1}{100}\\\\ \to \frac{1}{93.34} = \frac{700k + 1}{100} \\\\$

$\to 100 = 93.34(700k + 1) \\\\ \to 100 = 65,338k + 700 \\\\ \to 65,338k = -600 \\\\ \to k = \frac{-600}{ 65,338} \\\\ \to k= - 0.0091$

therefore plugging in the equation the above value of k

$\to \frac{1}{(100-2x)} = kt +\frac{1}{100} \\\\\to \frac{1}{(100-2x)} = -0.0091t + \frac{1}{100}\\\\\to \frac{1}{(100-2x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{2(50-x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{(50-x)} = \frac{1 -0.91t}{50}\\\\\to 50= (1-0.91t)(50-x)\\\\\to 50 = 50-45.5t-x-0.91tx\\\\\to x+0.91xt= -45.5t\\\\\to x(1+0.91t)= -45.5t\\\\\to x=\frac{-45.5t}{1+0.91t}$

Let y = C

, calculate C:

y = 3x

$y =3 \times \frac{-45.5t}{1+0.91t}$

amount of C formed in 28 mins

$x = \frac{-45.5t}{1+0.91t} ,$ plug t = 28

$\to x = \frac{-1274}{1+25.48} \\\\\to x = \frac{-1274}{26.48} \\\\\to x= -48.26$

therefore amount of C formed in 28 minutes is = 3x = 144.78 grams

C: $y =3 \times \frac{-45.5t}{1+0.91t}$

y= 136.5 =137

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lutik1710 [3]

Answer:

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Explanation:

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2. A 10.0 g ice cube, initially at 0.0ºC, is melted in 100.0 g of water that was initially 20.0ºC. After the ice has melted, the

arlik [135]

Answer:

The heat lost by the water $Q_{water} =$ 3.8 KJ

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The heat required to melt the ice $Q_{melt}$ = 3340 J

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Initial temperature of ice cube $T_{ice}$ = 0 °c

Mass of water $m_{water}$ = 100 gm

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⇒ $Q_{water} =$ 100 × 4.184 (20 - 10.93)

⇒ $Q_{water} =$ 3.8 KJ

This is the heat lost by the water.

(b). Heat gained by the ice cube $Q_{ice}$ = $m_{ice}$ $C_{ice}$ $(T_{f} - T_{ice} )$

⇒ $Q_{ice}$ = 10 × 2.093 × ( 10.93 - 0)

⇒ $Q_{ice}$ = 228.76 J

This is the heat gain by ice.

(C). Heat required to melt the ice $Q_{melt}$ = $m_{ice}$ × Latent Heat

⇒ $Q_{melt}$ = 10 × 334

⇒ $Q_{melt}$ = 3340 J

This is the heat required to melt the ice.

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