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zlopas [31]
3 years ago
14

7. Methane, CH, burns in oxygen gas to form water and carbon dioxide. Wequation for this reaction?a. CH4+ O H2O CO2b. CH4+ 40 2H

20+ CO2c. CH4 O2 H2O CO2d. CH4+ 202 2H20 CO2
8. How many atoms are present in 3 moles of chromium?c. 1.80 x 1024 atomsa. 6.02 x 1023 atomsb. 1.80 x 1023 atomsd. 52.00 atoms
9. How many grams of CO2 are in 2.1 mol of the compound?a. 21.0 gc. 66.0 gb. 44.0 gd. 92.4 g

Chemistry
1 answer:
Ainat [17]3 years ago
7 0
7) the answer is D.  You need to know that oxygen gas is O₂ which means that A and B can't be right.  Then you need to see which one is balanced and since D is and C is not, the correct answer has to be D.

8)  You need to know that 1 mole of anything contains 6.02×10²³ particles.  therefore you multiply <span>6.02×10²³ by 3 to get 1.806</span>×10²⁴ atoms.

9) You need to multiply the number of moles of carbon dioxide by the molar mass of carbon dioxide (44g/mol).  2.1mol×44g/mol=92.4g.  Therefore the answer is 92.4g of CO₂.

I hope this helps.  Let me know if anything is unclear.
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The distance between (–6, 2) and (8, 10) on a coordinate grid is 8.246
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9. Calculate the standard enthalpy of the 3rd reaction using the given data: A,H° = +52.96 kJ/mol AFH= -483.64 kJ/mol H2s)+I202
Mice21 [21]

Answer:

-586.56 kJ/mol is the standard enthalpy of the 3rd reaction.

Explanation:

H_2(g) +I_2(s) \rightarrow 2 HI(g) ,\Delta H^{o}_{1}= +52.96 kJ/mol...[1]

2 H_2(g) + O_2(g)\rightarrow 2 H_2O(g),\Delta H^{o}_{2}=-483.64 kJ/mol...[2]

4 HI(g)+O_2(g)\rightarrow 2 I_2(s)+2 H_2O(g) ,\Delta H^{o}_{3} =?..[3]

The unknown standard enthalpy of third reaction can be calculated by using Hess's law:

The law states that 'the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps'.

[2] - 2 × [1] = [3]

O_2+4HI\rightarrow 2H_2O(g)+2I_2(s)

\Delta H^{o}_{3}=\Delta H^{o}_{2}-2\times \Delta H^{o}_{1}

=-483.64 kJ/mol - 2\times (52.96 kJ/mol)=-586.56 kJ/mol

The standard enthalpy of the 3rd reaction is -586.56 kJ/mol.The negative sign indicates that energy is released during this reaction.

6 0
3 years ago
When metallic sodium is dissolved in liquid sodium chloride, electrons are released into the liquid. These dissolved electrons a
qaws [65]

Answer:

The edge of the length is \mathbf{L = 8.54 \times 10^{-10} \ m}

Explanation:

From the given information:

The associated energy for a particle in three - dimensional box can be expressed as:

E_n = \dfrac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2)

here;

h = planck's constant = 6.626 \times 10^{-34} \ Js

n_i = the quantum no in a specified direction

m = mass (of particle)

L = length of the box

At the ground state n_x = n_y = n_z=1

The energy at the ground state can be calculated by using the formula:

E_1 =\dfrac{3h^2}{8mL^2}

At first excited energy level, one of the quantum values will be 2 and the others will be 1.

Thus, the first excited energy will be: 2,1,1

∴

E_2 =\dfrac{(2^2+1^2+1^2)h^2}{8mL^2}

E_2 =\dfrac{(4+1+1)h^2}{8mL^2}

E_2 =\dfrac{(6)h^2}{8mL^2}

The transition energy needed to move from the ground to the excited state is:

\Delta E= E_2 - E_1

\Delta E= \dfrac{6h^2}{8mL^2}-  \dfrac{3h^2}{8mL^2}

\Delta E= \dfrac{3h^2}{8mL^2}} ----- (1)

Recall that:

the  wavelength identified with the electronic transition is: 800 nm

800 nm = 8.0  × 10⁻⁷ m

However, the energy-related with the electronic transition is:

\Delta E =\dfrac{hc}{\lambda}

\Delta E =\dfrac{6.626 \times 10^{-34} \times 2.99 \times 10^8}{8.0 \times 10^{-7} }

\Delta E =2.48 \times 10^{-19}  \ J

Replacing the value of \Delta E in (1); then:

2.48 \times 10^{-19}= \dfrac{3h^2}{8mL^2}}

Making the edge length L the subject of the formula; we have:

L = \sqrt{\dfrac{3h^2}{8m \times2.48 \times 10^{-19}} }

L = \sqrt{\dfrac{3\times (6.626 \times 10^{-34})^2}{8(9.1 \times 10^{-31} ) \times2.48 \times 10^{-19}} }

\mathbf{L = 8.54 \times 10^{-10} \ m}

Thus, the edge of the length is \mathbf{L = 8.54 \times 10^{-10} \ m}

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At a certain temperature a 42.55% (v/v) solution of ethyl alcohol, C2H5OH, in water has a density of 0.9027 g mL-1. The density
swat32

Answer:

a. 7.187M.

b. 7.962m.

c. 0.1846

Explanation:

This solution contains 42.55mL of ethyl alcohol in 100mL of solution.

a. Molar concentration = Moles ethyl alcohol / L solution:

<em>Moles ethyl alcohol -Molar mass: 46.07g/mol-</em>

42.55mL * (0.7782g/mL) * (1mol / 46.07g) = 0.7187 moles ethyl alcohol

<em>Liters solution:</em>

100mL * (1L / 100mL) = 0.100L

Molar concentration: 0.7187mol / 0.100L = 7.187M

b. Molal concentration: Moles ethyl alcohol (0.7187moles) / kg solution

<em>kg solution:</em>

100mL * (0.9027g / mL) * (1kg / 1000g) = 0.09027kg

Molal concentration: 0.7187mol / 0.09027kg = 7.962m

c. Mole fraction: Moles ethyl alcohol / Moles ethyl alcohol + moles water

<em>Moles water: -molar mass: 18.01g/mol-</em>

Volume water = 100mL - 42.55mL = 57.45mL

57.45mL * (0.9949g/mL) * (1mol/18.01g) = 3.1736 moles water

<em>Moles fraction:</em>

0.7187 moles ethyl alcohol / 0.7187 moles ethyl alcohol+3.1736 moles water

= 0.1846

4 0
3 years ago
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