Jade (a green gemstone) has a density of 3.6 g/mL. If a sample displaces 35.5 mL of water, what is the jade's mass?

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

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

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

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

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

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

7 0

1 year ago

Calculate the number of moles of each compound, given the number of molecules.

brilliants [131]

Answer:

1. $2.46\times 10^{21}$ molecules of CO₂

2. 10⁴ molecules of H₂O

3. 8.75×10³² molecules of C₆H₁₂O₆

Explanation:

1. $2.46\times 10^{21}$ molecules of CO₂

2. 10⁴ molecules of H₂O

3. 8.75×10³² molecules of C₆H₁₂O₆

5 0

2 years ago

The molarity of a solution prepared by dissolving 4.11 g of NaI in enough water to prepare 312 mL of solution is

Dimas [21]

Answer:

The correct answer is 8.79 × 10⁻² M.

Explanation:

Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,

No. of moles of NaI = Weight of NaI/ Molecular mass

= 4.11 / 149.89

= 0.027420

The vol. of the solution given is 312 ml or 0.312 L

The molarity can be determined by using the formula,

Molarity = No. of moles/ Volume of the solution in L

= 0.027420/0.312

= 0.0879 M or 8.79 × 10⁻² M

6 0

2 years ago

Una muestra de oxígeno ocupa 4.2 litros a 760 mm de hg . ¿ cual será el volumen del oxigeno a 415 mm de hg , si la temperatura p

ddd [48]

Teniendo en cuenta la ley de Boyle, el volumen del oxigeno a 415 mmHg, si la temperatura permanece constante, es 7,69 L.

Al aumentar el volumen, las partículas (átomos o moléculas) del gas tardan más en llegar a las paredes del recipiente y por lo tanto chocan menos veces por unidad de tiempo contra ellas. Esto significa que la presión será menor porque ésta representa la frecuencia de choques del gas contra las paredes.

De esta manera se relaciona la presión y el volumen, determinando la ley de Boyle que dice:

“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”

La ley de Boyle se expresa matemáticamente como:

P×V=k

Teniendo un estado inicial 1 y final 2, se cumple:

P₁×V₁= P₂×V₂

En este caso, se sabe que:

P₁= 760 mmHg
V₁= 4.2 L
P₂= 415 mmHg
V₂= ?

Reemplazando en la expresión matemática para la Ley de Boyle:

760 mmHg× 4.2 L= 415 mmHg× V₂

Resolviendo:

(760 mmHg× 4.2 L)÷ 415 mmHg= V₂

7,69 L= V₂

Finalmente, el volumen del oxigeno a 415 mmHg, si la temperatura permanece constante, es 7,69 L.

Aprende más:

5 0

1 year ago

Name the type of ion formed by the atom when it loses three electrons. Show by illustration

Nimfa-mama [501]

The atoms of elements can gain or lose electrons and become ions. Ions are charged particles that have gained or lost electrons. The atoms of elements can gain or lose electrons to form monatomic ions (made from a single atom of an element).

5 0

2 years ago

Read 2 more answers