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miv72 [106K]
3 years ago
9

In at least 4 sentences, explain how access to technology may have an impact to what energy sources are used? THINK- industriali

zed vs. non-industrialized nations.
Chemistry
1 answer:
larisa [96]3 years ago
8 0

Answer : attention swung away from renewable sources as the industrial revolution ... turbines have developed greatly in recent decades, solar photovoltaic technology is ... However, the variability of wind and solar power does not correspond with ... and 0.17 for solar PV, hence declared net capacity (DNC) is the figure

Explanation:

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Whag is the answer. please i need help​
icang [17]

Answer:

First option

Explanation:

Students 1 and 2 have precise measures yet is is not accurate since they measure around 5.

5 0
3 years ago
How many atoms are there in 0.028 grams<br> of iron metal?<br> 12:50 ✓
labwork [276]

Answer:

30100000000000000000 atoms

Explanation:

Ar of Fe : 56

0.028/56 = 0.0005 moles of Fe

0.0005 * 6.02 * 10^23(Avagardro's law of constant)

= 3.01*10^20 atoms

3 0
3 years ago
A sample of gas initially occupies 4.25 L at a pressure of 0.850 atm at 23.0°C. What will the volume be if the temperature is ch
Akimi4 [234]

Answer:

2.31\text{ L}

Explanation:

Here, we want to calculate the final volume

We use the general gas equation here:

\frac{P_1V_1}{T_1}\text{ = }\frac{P_2V_2}{T_2}

P1 is the initial pressure which is 0.850 atm

V1 is the initial volume which is 4.25 L

T1 is the initial temperature which is (23 + 273.15 = 296.15 K)

P2 is the final pressure which is 1.50 atm

V2 is the final volume which is unknown

T2 is the final temperature (11.5 + 273.15 = 284.65 K)

Substituting the values, we have:

\begin{gathered} \frac{0.850\text{ }\times4.25}{296.15}\text{ = }\frac{1.50\times V_2}{284.65} \\  \\ V_2\text{ = }\frac{284.65\text{ }\times0.850\times4.25}{1.5\times296.15} \\  \\ V_2\text{ = 2.31 L} \end{gathered}

6 0
1 year ago
Ion R2+ có phân lớp ngoài cùng là 3p6
amm1812
Ok and bro???? Like what even lol
4 0
3 years ago
if a lead can be extrated with 92.5% efficiency,what is the mass of ore required to make a lead sphere with 750%cm radius
Lady_Fox [76]
The mass of ore required is 21 700 t.

r = 750 cm

V = \frac{4}{3}  \pi  r^{3} = \frac{4}{3}  \pi  (750 cm)^{3} = 1.767 × 10⁹ cm³

The density of lead is 11.34 g/cm³.

So mass of lead sphere = 1.767 × 10⁹ cm³ × \frac{11.34 g}{1 cm^{3} } = 2.004 ×10¹⁰ g

2.004 ×10¹⁰ g × \frac{1 kg}{1000 g} = 2.004 × 10⁷ kg

2.004 × 10⁷ kg × \frac{1 t}{1000 kg} = 2.004 × 10⁴ t

92.5% efficiency means 92.5 t Pb per 100 t of ore.

Mass of ore = 2.004 × 10⁴ t Pb ×\frac{100 tore}{92.5 t Pb} = 2.17 × 10⁴ t ore = 21 700 t ore

6 0
3 years ago
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