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2 years ago

How are substances and mixtures different and alike???

1 answer:

6 0

A mixture is composed of more than one substance combined
A pure substance cannot be decomposed in any reaction
Both are considered types of substances
hope it helps you playa!

You might be interested in

Para formar bronce, se mezclan 150g de cobre a 1100°C y 35g de estaño a 560°C. Determine la temperatura final del sistema.

jek_recluse [69]

Answer:

La temperatura final del sistema es 1029,346 °C.

Explanation:

Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:

$m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)$

$(m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}$

$T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}}$ (1)

Donde:

$m_{Sn}$ - Masa del estaño, en gramos.

$m_{Cu}$ - Masa del cobre, en gramos.

$c_{Sn}$ - Calor específico del estaño, en calorías por gramo-grados Celsius.

$c_{Cu}$ - Calor específico del cobre, en calorías por gramo-grados Celsius.

$T_{Sn}$ - Temperatura inicial del estaño, en grados Celsius.

$T_{Cu}$ - Temperatura inicial del cobre, en grados Celsius.

Si sabemos que $m_{Cu} = 150\,g$ , $m_{Sn} = 35\,g$ , $c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}$ , $c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}$ , $T_{Sn} = 560\,^{\circ}C$ y $T_{Cu} = 1100\,^{\circ}C$ , entonces la temperatura final del sistema es:

$T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}$

$T = 1029,346\,^{\circ}C$

La temperatura final del sistema es 1029,346 °C.

3 0

2 years ago

What is the predicted change in the boiling point of water when 1.50 g of

dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

⠀

$\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}$

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The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.

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$\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}$

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Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

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'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

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'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

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To find molality, we have to divide no. of moles of solute by weight of solution

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While first we need to no. of moles

$\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072$

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Now, we will find molality

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$\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}$

⠀

$\textsf{ \large{ \underline{Now substituting the required values}}}$

⠀

$\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}$

⠀

Henceforth, the change in boiling point is 0.00735°C.

7 0

1 year ago

PLZ HELP! How do you think changing the angle of a ramp will affect work done?

kogti [31]

Answer: The steepness of a ramp affects it by making it easier or harder.

Explanation: It's a bit situational. If you were going up a steep ramp with a heavy load, it will increase the work necessary, whereas if you were going down a ramp, it would decrease the work necessary. If you need this simply put, think about biking up and down a hill. It would be easier going down than up.

6 0

2 years ago

4. How might the process of making paper from wood be changed to produce paper that is not acidic?

kicyunya [14]

Answer:it is sterilized in the process

Explanation:

5 0

2 years ago

Decreasing the temperature of the reaction 3H2 + N2 2NH3. In this reaction, the product absorbs heat. WHICH WAY WILL THE REACTIO

Alex

Answer:

the reaction will shift towards the “heat”—shifts to the left

Explanation:

To summarize:

o If temperature increases (adding heat), the reaction will shift away from the “heat” term and go in the

endothermic direction.

o If temperature decreases (removing heat), the reaction will shift towards the “heat” term and go in the

exothermic direction.

o NOTE: The endothermic direction is always away from the “heat” term and the exothermic direction is

towards the “heat” term.

Therefore the reaction will shift towards the “heat”—shifts to the left

3 0

3 years ago