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natulia [17]
3 years ago
5

How many hours are in 188 years

Chemistry
1 answer:
AURORKA [14]3 years ago
7 0

Answer: about 1,705,536 hrs

24 hrs = 1 day

189 hrs = 1 week

756 = 1 month

9072 = 1 yr

1,705,536 = 188 yrs

Explanation:

Trust Milky. Milky smart

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The question is below
rodikova [14]
The anode is the electrode where the oxidation occurs.

Cathode is the electrode where the reducction occurs.

Equations:

  Mn(2+) + 2e-  --->  Mn(s)                      Eo = - 1.18 V
2Fe(3+) + 2e-  ----> 2 Fe(2+)                2Eo = + 1.54 V

The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.

Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72

Answer: 2.72 V

3 0
3 years ago
Assuming that gasoline is 100% isooctane, that isooctane burns to produce only CO2CO2 and H2OH2O, and that the density of isooct
Aleksandr [31]

Answer:

1.12×10¹¹ kg of CO₂ are produced with 4.6×10¹⁰ L of isooctane

Explanation:

Let's state the combustion reaction:

C₈H₁₈  +  25/2O₂  →   8CO₂  +  9H₂O

Let's calculate the mass of isooctane that reacts.

Density = Mass / Volume

Density . Volume = Mass

First of all, let's convert the volume in L to mL, so we can use density.

4.6×10¹⁰ L . 1000 mL / 1L = 4.6×10¹³ mL

0.792 g/mL . 4.6×10¹³ mL = 3.64 ×10¹³ g

This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction

3.64 ×10¹³ g . 1mol / 114 g = 3.19×10¹¹ mol

Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide

Therefore 3.19×10¹¹ mol would produce (3.19×10¹¹ mol . 8)  = 2.55×10¹² moles of CO₂

Now, we can determine the mass of produced CO₂ by multipling:

moles . molar mass

2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g of CO₂

If we convert to kg  1.12×10¹⁴ g / 1000 =  1.12×10¹¹ kg

6 0
3 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
3 years ago
Rank the compounds in each set in order of increasing acid strength.
Viktor [21]

Answer:

See explanation

Explanation:

For this question, we have to remember the effect of an atom with high <u>electronegativity</u> as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an <u>inductive effect</u>. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be <u>weaker</u> and the compound will be more acid (because is easier to produce the hydronium ion H^+).

With this in mind, for A in the last compound, we have <u>2 Br atoms</u> near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have <u>more acidity</u>. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.

In B, the difference between the molecules is the <u>position</u> of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a <u>higher inductive effect</u> and more <u>acidity</u>.

See figure 1

I hope it helps!

5 0
3 years ago
The reactants of two chemical equations are listed.
hjlf
The correct answer is b
8 0
3 years ago
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