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3 years ago

Identify the major ionic species present in an aqueous solution of C12H22O11 (sucrose).

Chemistry

1 answer:

Marina86 [1]3 years ago

4 0

<h3>Answer;</h3>

No ions present

<h3>Explanation;</h3>

Ionic compounds are compounds made up of ions. These ions are atoms that gain or lose electrons, giving them a net positive or negative charge.
Atoms that gain electrons and therefore have a net negative charge are known as anions. Conversely, atoms that lose electrons have a net positive charge are called cations.
C12H22O11 (sucrose) is not an ionic compound, and therefore does not have any ions. Sucrose is a molecular compound.

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Joseph divides 8.64 by 2.0. How many significant figures should his answer have?

nadezda [96]

Answer:

2 significant figures

Good luck with your work! :)

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3 years ago

Which of the following best describes atoms? *

Masja [62]

a. Smallest basic units of matter

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Molecule: A molecule is a substance which exists in free state but doesn't take part in chemical reactions.

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3 years ago

How many atoms of carbon are found in a 30.0 gram sample of the element?

saveliy_v [14]

Answer:

1.50 × 10²⁴ atoms C

Explanation:

Step 1: Define

Molar mass of C - 12.01 g/mol

Avagadro's Number: 6.02 × 10²³ atoms, molecules, formula units, etc.

Step 2: Use Dimensional Analysis

$30.0 \hspace{3} g \hspace{3} C(\frac{1 \hspace{3} mol \hspace{3} C}{12.01 \hspace{3} g \hspace{3} C} )(\frac{6.02(10)^{23} \hspace{3} atoms \hspace{3} C}{1 \hspace{3} mol \hspace{3} C} )$ = 1.50375 × 10²⁴ atoms C

Step 3: Simplify

We are given 3 sig figs.

1.50375 × 10²⁴ atoms C ≈ 1.50 × 10²⁴ atoms C

3 0

3 years ago

Write a balanced equation for the combustion of C7H16(l) (heptane) -- i.e. its reaction with O2(g) forming the products CO2(g) a

JulsSmile [24]

Answer:

The standard enthalpy of reaction = -4854.7kJ

The difference: ΔH-ΔE = Δ(PV) = Δn.R.T = 9910.288 J ≈ 9.91 kJ

Explanation:

The balanced chemical equation for the combustion of heptane:

C₇H₁₆ (l) + 11 O₂ (g) → 7 CO₂ (g) + 8 H₂O (l)

Given: The standard enthalpy of formation ( $\Delta H _{f}^{\circ }$ ) for: C₇H₁₆ (l) = -187.8 kJ/mol, O₂ (g) = 0 kJ/mol, CO₂ (g) = -393.5 kJ/mol, H₂O (l) = -286 kJ/mol

To calculate the standard enthalpy of reaction ( $\Delta H _{r}^{\circ }$ ) can be calculated by the Hess's law:

$\Delta H _{r}^{\circ } = \left [\sum \nu \cdot\Delta H _{f}^{\circ }(products) \right ] - \left [\sum \nu\cdot\Delta H _{f}^{\circ }(reactants) \right ]$

Here, $\nu$ is the stoichiometric coefficient

⇒ $\Delta H _{r}^{\circ } =$

$\left [ 7\times \Delta H _{f}^{\circ }\left (CO_{2}\right )+ 8\times \Delta H _{f}^{\circ }\left (H_{2}O \right )\right ]$

$- \left [1\times \Delta H _{f}^{\circ }\left (C_{7}H_{16}\right ) +11\times \Delta H _{f}^{\circ }\left (O_{2} \right ) \right ]$

$=\left [ 7\times \left (-393.5 kJ/mol \right )+ 8\times \left (-286 kJ/mol \right )\right ]$

$-\left [1\times \left (-187.8 kJ/mol \right ) +11\times \left (0 kJ/mol \right ) \right ]$

⇒ $\Delta H _{r}^{\circ } = \left [ \left (-2754.5 \right )+ \left (-2288 \right )\right ]\left -[ \left (-187.8 \right ) +\left (0 \right )\right ]$

⇒ $\Delta H _{r}^{\circ } = \left [ -5042.5 ]\left -[ -187.8] = \left ( -4854.7kJ \right )$

To calculate the difference: ΔH-ΔE=Δ(PV)

We use the ideal gas equation: P.V = n.R.T

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T

Given: Temperature:T = 298K, R = 8.314 J⋅K⁻¹⋅mol⁻¹

Δn = number of moles of gaseous products - number of moles of gaseous reactants = (7)- (11) = (-4)

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T = (-4 mol) × (8.314 J⋅K⁻¹⋅mol⁻¹) × (298K) = 9910.288 J = 9.91 kJ (∵ 1 kJ = 1000J )

8 0

3 years ago

The radioisotope radon-222 has a half life of 3.8 days. How much of a 10 gram sample of radon-222 would be left after 15.2 days?

djyliett [7]

Answer:

0.625 g

Explanation:

Given data:

Half life of radon-222 = 3.8 days

Total mass of sample = 10 g

Mass left after 15.2 days = ?

Solution:

Number of half lives = T elapsed / Half life

Number of half lives = 15.2 / 3.8

Number of half lives = 4

At time zero = 10 g

At first half life = 10 g/2 = 5 g

At 2nd half life = 5 g/ 2= 2.5 g

At third half life = 2.5 g/2 = 1.25 g

At 4th half life = 1.25 g/2 = 0.625 g

4 0

3 years ago