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kodGreya [7K]
3 years ago
8

Why is fusion not used to generate electrical power ?

Chemistry
2 answers:
Goryan [66]3 years ago
7 0

Answer:

Fusion doesn't produce runway chain reactions the way fission can, so there's no need to worry about meltdowns.

salantis [7]3 years ago
4 0
Way too much energy
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How many mL of 2.0 M KOH are necessary to neutralize 50 mL of 1 M HCl?
worty [1.4K]

Answer:

25mL

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this can help you. Have a nice day!

6 0
4 years ago
PLEASE HELP IT DUE TODAY WILL GIVE 100 BRAINLY POINTS AND BRAINLYEST
inessss [21]

Answer:

1. The energy intensity from the ruler going through the air

2. This one is pretty simple, you can add or take away things. This question is just asking you to list sounds from high pitch, to lowest pitch. Here are mine:

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Explanation:

4 0
3 years ago
How many grams of Pb are contained in a mixture of 0.135 kg each of PbCl(OH) and Pb2Cl2CO3?
Dahasolnce [82]

Hey there!:

Given the mass of PbCl(OH) :

0.135 Kg = 0.135 Kg*(1000g / 1Kg)  = 135 g

Molecular mass of PbCl(OH)  = 207+35.5+16+1  = 259.5 g / mol

Atomic mass of Pb = 207 g/mol

Hence mass of Pb in 135 g  PbCl(OH)  :

(207 g Pb /  259.5 g PbClOH) * 135g PbClOH  =

0.79768 * 135 =>  107.68 g of Pb

For Pb2Cl2CO3  :

Given the mass of Pb2Cl2CO3  :

0.135 Kg = 0.135 Kgx(1000g / 1Kg)  = 135 g

Molecular mass of Pb2Cl2CO3  = 2*207+2*35.5+12+3*16  = 545 g / mol

Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3  = 2*207 = 414 g

Hence mass of Pb in 135 g  Pb2Cl2CO3:

(414 g Pb /  545 g PbClOH) * 135g PbClOH  =

0.75963 * 135 => 102.55 g of Pb2Cl2CO3


Hope that helps!

8 0
3 years ago
Read 2 more answers
Please help me as soon as possible! Please help me as soon as possible!
Nastasia [14]

The Answer is D. Suspending a heavy weight with a strong chain.

6 0
4 years ago
What is the solution's freezing point: 15 g of CH4N2O (Molar mass = 60.055 g/mol) in 200. g of H2O? (Kf = 1.86 (°C·kg)/mol)
AURORKA [14]

Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :

1 mol of urea =15/60.055 = 0.25mol

therefore 200g of water contain 0.25mol

the next step is to determine the malality of our solution in 200g of water, to do this we say:

200 g = 1Kg/1000g = 0.2kg

therefor 0.25mol/0.2Kg = 1.25mol/kg

and from the equation:

we know that i = 1

we are given Kf

b is the molality that we just calculated

therefore;

the solutions freezing point is -2.325°C

8 0
1 year ago
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