1)

One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will

saw5 [17]

Explanation:

The given data is as follows.

Boiling point of water ( $T^{o}_{b}) = 100^{o}C$ = (100 + 273) K = 323 K,

Boiling point of solution ( $T_{b}) = 101.24^{o}C$ = (101.24 + 273) K = 374.24 K

Hence, change in temperature will be calculated as follows.

$\Delta T_{b} = (T_{b} - T^{o}_{b})$

= 374.24 K - 323 K

= 1.24 K

As molality is defined as the moles of solute present in kg of solvent.

Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

Let molar mass of the solute is x grams.

Therefore, Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

m = $\frac{288 g \times 1000}{x g \times 90}$

= $\frac{3200}{x}$

As, $\Delta T_{b} = k_{b} \times molality$

$1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}$

x = $\frac{0.512 ^{o}C/m \times 3200}{1.24}$

= 1321.29 g

This means that the molar mass of the given compound is 1321.29 g.

It is given that molecular formula is $C_{n}H_{2n}O_{n}$ .

As, its empirical formula is $CH_{2}O$ and mass is 30 g/mol. Hence, calculate the value of n as follows.

n = $\frac{\text{Molecular mass}}{\text{Empirical mass}}$

= $\frac{1321.29 g}{30 g/mol}$

= 44 mol

Thus, we can conclude that the formula of given material is $C_{44}H_{88}O_{44}$ .

4 0

3 years ago

As you move upward, from level, in an energy pyramid, available energy ____.

svetlana [45]

The available energy decreases as one moves upward in an energy pyramid.

<h3>Energy pyramid</h3>

The energy pyramid represents a model of how energy is transferred from one trophic level to another in ecosystems.

Energy is transferred from producers to primary consumers, from primary to secondary consumers, from secondary to tertiary consumers, and so on.

Only about 10% of the available energy in one trophic level is transferred to the next with the remaining 90% being lost as heat to the environment.

Thus, as one moves up the energy pyramid, the available energy decreases. This is why organisms at the higher end of the energy pyramid have to devise an efficient way of extracting energy from their foods.

More on energy pyramid can be found here: brainly.com/question/2515928

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8 0

2 years ago

______________________ produce evidence that helps to answer important questions or could also lead to new questions. A.investig

siniylev [52]

Answer:

B

Explanation:

6 0

3 years ago

Read 2 more answers

Cr2o2−7(aq)+i−(aq)→cr3+(aq)+io−3(aq) (acidic solution) express your answer as a chemical equation. identify all of the phases in

Rashid [163]

<span>Answer: Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either. Start with your 2 half reactions: I- --> IO3- Cr2O72- --> 2 Cr3+ Balance O by adding H2O: I- + 3 H2O --> IO3- Cr2O72- --> 2 Cr3+ + 7H2O Balance H by adding H+: I- + 3 H2O --> IO3- + 6 H+ Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O Balance charge by adding e-: I- + 3 H2O --> IO3- + 6 H+ + 6 e- Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give: Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>

4 0

3 years ago

In the following reaction, H3PO4 (aq) + H2O (l) ⇄ H2PO4– (aq) + H3O+ (aq) what happens when more H2PO4– (aq) is added to the sol

kirza4 [7]

Answer:

The concentration of H₃PO₄ will increase.

Explanation:

H₃PO₄(aq) + H₂O(l) ⇄ H₂PO₄⁻(aq) + H₃O⁺(aq)

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

If we add more H₂PO₄⁻, the position of equilibrium will move to the left to get rid of the added H₂PO₄⁻.

The concentration of H₃PO₄ will increase.

7 0

3 years ago