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2 years ago

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Chemistry

1 answer:

ddd [48]2 years ago

6 0

Explanation:

Uranium-238 undergoes a radioactive decay series consisting of 14 separate steps before producing stable lead-206. This series consists of eight α decays and six β decays.

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Mrs. Rushing fills a balloon with hydrogen gas to demonstrate its ability to burn. Which combination could she

Mars2501 [29]

Answer:

H2SO, and CaCO

Explanation:

8 0

3 years ago

Read 2 more answers

Please help, with step by step work

natali 33 [55]

$\qquad$ ☀️ $\pink{\bf{ {Answer = \: \: 85.57g }}}$

Molar mass of $\bf Cu_2O$

$\qquad$ $\twoheadrightarrow\sf 63.546 \times 2 +16$

$\qquad$ $\pink{\twoheadrightarrow\bf 143.092 g}$

As we know–

1 mol = $\bf 6.02×10^{23}$ formula units

1 mol $\bf Cu_2O$ = 143.092 g = $\bf 6.02×10^{23}$ formula units

Henceforth –

$\bf 3.60×10^{23}$ formula units $\bf Cu_2O$ –

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×10^{23 }}{6.02×10^{23}}$

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23 }}}{6.02×\cancel{10^{23}}}$

$\qquad$ $\pink{:\implies\bf 85.57 g}$

5 0

2 years ago

Calculate the standard entropy change for the following reactions at 25°C.

Art [367]

Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.

(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)

Sº 159.9 205.2 213.8 188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K

ΔSº = 216.10 J/K

(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)

Sº 151.0 205.2 213.8 248.2

(J/Kmol)

ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

Sº 173.3 205.2 213.8 188.8

(J/Kmol)

ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.

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3 years ago

A comic book villain is holding you at gun point and is making you drink a sample of acid he gives you a beaker with 100 ml of a

miss Akunina [59]

Answer:

b. 10 mL

Explanation:

First we calculate the amount of H⁺ moles in the acid:

pH = -log [H⁺]

[H⁺] = $10^{-pH}$

[H⁺] = 10⁻⁵ = 1x10⁻⁵M

100 mL ⇒ 100 / 1000 = 0.100 L

1x10⁻⁵M * 0.100 L = 1x10⁻⁶ mol H⁺

In order to have a neutral solution we would need the same amount of OH⁻ moles.

We can use the pOH value of the strong base:

pOH = 14 - pH

pOH = 14 - 10 = 4

Then we calculate the molar concentration of the OH⁻ species in the basic solution:

pOH = -log [OH⁻]

[OH⁻] = $10^{-pOH}$ = 1x10⁻⁴ M

If we use 10 mL of the basic solution the number of OH⁻ would be:

10 mL ⇒ 10 / 1000 = 0.010 L

1x10⁻⁴ M * 0.010 L = 1x10⁻⁶ mol OH⁻

It would be equal to the moles of H⁺ so the answer is b.

7 0

2 years ago

Read 2 more answers

How much faster will ammonia vapor travel across a room then carbon dioxide

FromTheMoon [43]

Answer:

D) 1.61 times faster

Explanation:

$vrms_{NH3}$ = √(3)RTM

R constant= 0.08206

T=constant, so in this problem we dont need a value for it

M=17.031 g/mol

√(3)(0.08206)(17.031)= 2.047

$vrms_{CO2}$ = √(3)RTM

R constant= 0.08206

T=constant, so in this problem we dont need a value for it

M= 44.01 g/mol

√(3)(0.08206)(44.01)= 3.29

Since we are trying to measure how much faster NH3 will be, we have to realize that mass and speed have an inverse relationship.

So instead of doing (2.047)/(3.29) = 6.22

we have to flip the values to get (3.29)/(2.047)= 1.61

6 0

2 years ago