The shape of the nucleus is maintained by

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

2 years ago

Plz download the doc and answer its due in 20 mins like bro wth

4vir4ik [10]

Answer:

and how do we know this is not a virus

Explanation:

sorry dude im just saying

4 0

3 years ago

Read 2 more answers

Which formula is an empirical formula?

poizon [28]

<h2>Hello!</h2>

The answer is:

The empirical formula is the option B. $NH_{3}$

The empirical formula of a compound is the simplest formula that can be written. On the opposite, the molecular formula involves a variant of the same compound, but it can be also simplified to an empirical formula.

$MolecularFormula=n(EmpiricalFormula)$

We are looking for a formula that cannot be simplified by dividing the number of molecules/atoms that conforms the compound.

Let's discard option by option in order to find which formula is an empirical formula (cannot be simplified)

A. $N_{2}O_{4}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$N_{2}O_{4}=2(NO_{2})$

B. $NH_{3}$

It's an empirical formula since it cannot be obtained by the multiplication of a whole number and the simplest formula. It's the simplest formula that we can find of the compound.

C. $C_{3}H_{6}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$C_{3}H_{6}=3(CH_{2})$

D. $P_{4}O_{10}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$P_{4}O_{10}=2(P_{2}O_{5})$

Hence, the empirical formula is the option B. $NH_{3}$

Have a nice day!

6 0

3 years ago

Read 2 more answers

For each element, predict where the "jump " occurs for successive ionization energies. (For example, does the jump occur between

vichka [17]

Answer:

A jump occurs when a core electron is removed.

Explanation:

A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.

For Beryllium, the electronic configuration of is 1s2 2s2.

There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron

The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron

The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron

The electronic configuration of Lithium is 1s2 2s1

There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.

8 0

4 years ago

A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution? 1.0 × 10–5 M OH– 1.0 ×

Triss [41]

Answer:

1.0 × 10⁻⁹ M.

Explanation:

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

[H₃O⁺] = 1.0 x 10⁻⁵ M.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(1.0 x 10⁻⁵ M) = 1.0 × 10⁻⁹ M.

6 0

3 years ago

Read 2 more answers