What is the volume of SO3 in 835 g SO3?

there are 1540 megawatts of wind-generated electricity produced globally every year. This amount is equivalent to how many kilow

Reptile [31]

1.54 x 10⁶kW

Explanation:

The problem here is converting from megawatts into kilowatts.

Given 1540megawatts.

The mega- and kilo- watts are prefixes that denotes multiples of units.

1 megawatt = 10⁶watts

1 kilowatt = 10³watts

Now given 1540megawatts power to Kilowatt power;

We see that :

1000kW = 1mW

1540mW x $\frac{1000kW}{1mW}$ = 1.54 x 10⁶kW

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4 0

3 years ago

Which of the following keeps satellite in orbit

BabaBlast [244]

Answer:

Gravity

Explanation:

Without gravity it would just float in space.

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4 0

3 years ago

Which of the following is an example of eukaryotic cells? animals bacteria archaea

ElenaW [278]

Answer:

Animals

Explanation:

Eukaryotic organisms are multi-celled. Animals and plants are a major example of that

8 0

3 years ago

What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0

expeople1 [14]

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = 45.1mL

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = $\frac{[OH^-][HNO_2]}{[NO_2]}$

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

Where X is reaction coordinate

Replacing in Kb:

1.41x10⁻¹¹ = $\frac{[X][X]}{[0.211 -X]}$

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ FALSE ANSWER. There is no negative concentrations.

X = 1.72x10⁻⁶. Right answer.

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; pH = 8.24

3 0

3 years ago

The radioactive substance cesium-137 has a half-life of 30 years. The amount At (in grams) of a sample of cesium-137 remaining a

stealth61 [152]

Answer: The amount of sample left after 20 years is 288.522 g and after 50 years is 144.26 g

Explanation:

We are given a function that calculates the amount of sample remaining after 't' years, which is:

$A_t(t)=458\times (\frac{1}{2})^{\frac{t}{30}$

For t = 20 years

Putting values in above equation:

$A_t(t)=458\times (\frac{1}{2})^{\frac{20}{30}$

$A_t(t)=288.522g$

Hence, the amount of sample left after 20 years is 288.522 g

For t = 50 years

Putting values in above equation:

$A_t(t)=458\times (\frac{1}{2})^{\frac{50}{30}$

$A_t(t)=144.26g$

Hence, the amount of sample left after 50 years is 144.26 g

6 0

4 years ago