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2 years ago

13

Provided is a list of five different substances. How many of these substances are elements?.

1 answer:

Lina20 [59]2 years ago

6 0

Answer:

118

Explanation:

Of these 118 elements, 94 occur naturally on Earth. Six of these occur in extreme trace quantities: technetium, atomic number 43; promethium, number 61; astatine, number 85; francium, number 87; neptunium, number 93; and plutonium, number 94.

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Please tell me answers I am given you brainliest

Zina [86]

Answer:

1 = one

2 = fluorine

3 = Bromine

4 = increase

5 = first increaseas and then decreases

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3 years ago

Which of the following would be a result of increased solar activity?

scoundrel [369]

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3 years ago

Read 2 more answers

The temperature of a freezer started at 18şc. After cooling for a few hours, the freezer had a temperature of –12şc. What is the

andre [41]

20 minus 2 plus 1 equals negative -2000000

8 0

2 years ago

If 1.27 moles of bromine gas (Br2) react with excess phosphorus (P), how many moles of phosphorus tribromide (PBr3) will be prod

frez [133]

Answer:

0.85 mole of PBr3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Br2 + 2P —> 2PBr3

From the balanced equation above,

3 moles of Br2 reacted to produce 2 moles of PBr3.

Therefore, 1.27 moles of Br2 will react to produce = (1.27 x 2)/ 3 = 0.85 mole of PBr3.

Therefore, 0.85 mole of PBr3 is produced by the reaction.

8 0

3 years ago

In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in

RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 $m^{3}$

Density of water = 1000 $kg/m^{3}$

Therefore, mass of water = Density × Volume

= $1000 kg/m^{3} \times 0.25 m^{3}$

= 250 kg

Initial Temperature of water ( $T_{1}$ ) = $20^{o}C$

Final temperature of water = $140^{o}C$

Heat of vaporization of water ( $dH_{v}$ ) at $140^{o}C$ is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point ( $140^{o}C$ ) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

= 62.5 (kg) × 2133 (kJ/kg)

= $133.3 \times 10^{3}$ kJ

All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = $\frac{133.3 \times 10^{3}kJ}{3600 s}$ = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

= 125520 kJ

Time required = Heat required / Power rating

= $\frac{125520}{37}$

= 3392 sec

Time required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 0.25 $m^{3}$ water is calculated as follows.

$\frac{3392 sec}{60 sec/min}$

= 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0

3 years ago