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3 years ago

What are advantages of using uranium as an energy source

2 answers:

7 0

-Small amounts of Uranium generate large amounts of energy
-Uranium occurs in huge reserves
-It has a longer lifetime than other non-renewable sources of energy

MariettaO [177]3 years ago

4 0

Question:
What are advantages of using Uranium as an energy source?

Answer(s):
-Small amounts of Uranium generate large amounts of energy
-Uranium occurs in huge reserves
-It has a longer lifetime than other non-renewable sources of energy

-Brainly Answerer

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HEllo i need help with reading how do i read

hram777 [196]

Try to sound out the words ! :)

5 0

3 years ago

Given the following reaction: 3CuCl2(aq) 2Na3PO4(aq) --> Cu3(PO4)2(s) 6NaCl(aq) MM (g/mol) 134.45 163.94 380.58 58.44 If 285

nata0808 [166]

Answer:

227.78g of the precipitate are produced

Explanation:

Based on the reaction, 3 moles of CuCl2 produce 1 mole of Cu3(PO4)2 (The precipitate).

To solve this question we need to find the moles of CuCl2 added. With these moles and the reactio we can find the moles of Cu3(PO4)2 and its mass as follows:

Moles CuCl2:

285mL = 0.285L * (6.3mol / L) = 1.7955 moles CuCl2

Moles Cu3(PO4)2:

1.7955 moles CuCl2 * (1mol Cu3(PO4)2 / 3mol CuCl2) = 0.5985 moles Cu3(PO4)2

Mass Cu3(PO4)2 -380.58g/mol-

0.5985 moles Cu3(PO4)2 * (380.58g/mol) =

227.78g of the precipitate are produced

7 0

3 years ago

Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, y

denis23 [38]

Answer:

Volume of $CHCl_3$ = 15.31 mL

Volume of $CHBr_3$ = 4.69 mL

Explanation:

Given that:

the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of $CHCl_3$ = 1.492 g/mL

The density of $CHBr_3$ = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of $CHCl_3$ = P ml

and the volume of $CHBr_3$ = Q ml

the sum of their volumes should be equal to the total volume of the mixture

$P \ ml + Q \ ml = 20 ml$ ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

$\mathtt{(Density \ of \ CHCl_3 \times Vol. \ of \ CHCl_3 ) + (Density \ of \ CHBr_3 \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density \ of \ mixture \times volume \ of \ the \ mixture)}$

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g = 36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of $CHCl_3$ = 15.31 mL

Volume of $CHBr_3$ = 4.69 mL

6 0

3 years ago

Perform the following calculations and give your answer with the correct number of significant figures:

GarryVolchara [31]

I think this is the answer try it

172.3995

4 0

3 years ago

Read 2 more answers

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago