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tatuchka [14]
3 years ago
12

Which is the fastest in the universe apart from the light

Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
3 0
Well i did some research and i found out now im not sure if this is completely true but supposedly tachyons are even faster than the speed of light
You might be interested in
Which element decreases its oxidation number in this reaction? bicl2 + na2so4 → 2nacl + biso4?
kifflom [539]
The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction 
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl 
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.

5 0
3 years ago
Read 2 more answers
B) Why is aluminium extracted using electrolysis and not carbon?
Elena-2011 [213]
The extraction is done by electrolysis. The ions in the aluminium oxide must be free to move so that electricity can pass through it. Aluminium oxide has a very high melting point (over 2000°C) so it would be expensive to melt it. The use of cryolite reduces some of the energy costs involved in extracting aluminium.
6 0
3 years ago
Read 2 more answers
Calculate the molarity of 90.0 mL of a solution that is 0.92 % by mass NaCl. Assume the density of the solution is the same as p
jeka94
Answer is: molarity is 0,155 M.
V(solution) = 90,0 mL = 0,09 L.
ω(NaCl) = 0,92% ÷ 100% = 0,0092.
d(solution) = 1 g/mL.
m(solution) = V(solution) · d(solution).
m(solution) = 90 mL · 1 g/mL = 90 g.
m(NaCl) = 90 g · 0,0092 = 0,828 g.
n(NaCl) = 0,828 g ÷ 58,4 g/mol.
n(NaCl) = 0,014 mol.
c(solution) = 0,014 mol ÷ 0,09 L.
c(solution) = 0,155 mol/L.

8 0
3 years ago
The second-order rate constant for the following gas-phase reaction is 0.041 1/MLaTeX: \cdotâs. We start with 0.438 mol C2F4 in
pantera1 [17]

Answer:

134.8 seconds is the half-life (in seconds) of the reaction for the initial C_2F_4 concentration

Explanation:

Half life for second order kinetics is given by:

t_{\frac{1}{2}=\frac{1}{k\times a_0}

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

t_{\frac{1}{2} = half life

k = rate constant

a_0 = initial concentration

a = Final concentration of reactant after time t

We have :

C_2F_4 \longrightarrow \frac{1}{2} C_4F_8

Initial concentration of C_2F_4=[a_o]=\frac{0.438 mol}{2.42 L}=0.1810 mol/L

Rate constant = k = 0.041 M^{-1} s^{-1}

t_{\frac{1}{2}=\frac{1}{k\times a_0}

=\frac{1}{0.041 M^{-1} s^{-1}\times 0.1810 mol/L}

t_{1/2}=134.8 s

134.8 seconds is the half-life (in seconds) of the reaction for the initial C_2F_4 concentration

3 0
3 years ago
A 3.0-mol sample of kclo3 was decomposed according to the equation 2kclo3(s) --> 2kcl(s) + 3o2(g) how many moles of o2 are fo
d1i1m1o1n [39]
2moles of O2 are formed.
5 0
3 years ago
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