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ikadub [295]
3 years ago
13

Emits huge quantities of light and heat due to the nuclear

Chemistry
1 answer:
kifflom [539]3 years ago
8 0
Black hole I think that is right
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What is the number assigned to an element on the Periodic Table?
spayn [35]

Answer:

Element Atomic Number One number you will find on all periodic tables is the atomic number for each element.

Explanation:

<em>hope it helps :))</em>

6 0
3 years ago
What volume of 0.500 m h2so4 is needed to react completely with 20.0 ml of 0.400 m lioh?
ExtremeBDS [4]
The balanced chemical equation for the above reaction is as follows;
2LiOH + H₂SO₄ ---> Li₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
Number of OH⁻ moles reacted = number of H⁺ moles reacted at neutralisation
Number of LiOH moles reacted = 0.400 M / 1000 mL/L x 20.0 mL = 0.008 mol 
number of H₂SO₄ moles reacted - 0.008 mol /2 = 0.004 mol 
Number of H₂SO₄ moles in 1 L - 0.500 M
This means that 0.500 mol  in 1 L solution 
Therefore 0.004 mol in - 1/0.500 x 0.004 = 0.008 L 
therefore volume of acid required = 8 mL 
4 0
3 years ago
Which molecule is a stereoisomer of trans-2-butene?
S_A_V [24]

Answer:

I can't say that it is definitely write.

HHH

H-C-C-C

6 0
2 years ago
The formula equation of Acetylene + oxygen ----&gt; carbon dioxide + water ​
snow_lady [41]

Answer:

The final balanced equation is : 2C2H2+5O2→4CO2+2H2O.

8 0
3 years ago
It is found that a gas undergoes a first-order decomposition reaction. If the rate constant for this reaction is 8.1 x 10-2 /min
kap26 [50]

Answer:

28.43 min

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 8.1\times 10^{-2} min⁻¹

Initial concentration [A_0] = 0.1 M

Final concentration [A_t] = 1.0\times 10^{-2} M

Time = ?

Applying in the above equation, we get that:-

1.0\times 10^{-2}=0.1e^{-8.1\times 10^{-2}\times t}

0.1e^{-8.1\times \:10^{-2}t}=10^{-2}

e^{-8.1\times \:10^{-2}t}=\frac{1}{10}

\ln \left(e^{-8.1\times \:10^{-2}t}\right)=\ln \left(\frac{1}{10}\right)

t=28.43\ min

3 0
4 years ago
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