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maxonik [38]
2 years ago
7

Which types of particles are involved in a fission reaction?

Chemistry
2 answers:
mihalych1998 [28]2 years ago
5 0
The first one a neutral neutron strikes a large neutral nucleus
kati45 [8]2 years ago
3 0

Answer:

c sry if wrong

Explanation:

You might be interested in
Which pair of elements can form an ionic compound? A) Two atoms of hydrogen B) One atom of sodium and one atom of fluorine C) On
polet [3.4K]
The answer would be B. <span>One atom of sodium and one atom of fluorine</span>
5 0
2 years ago
Read 2 more answers
C4H10 + 02 ➡️ H20 + CO2<br> balance the equation
Alexeev081 [22]

(Can I have Brainlist)?

Answer:

2C4H10 + 13O2 —> 8CO2 + 10H2O. Oxidation reaction

8 (4 moles CO2 per mole butane)

Explanation:

could be written C4H10 + 6 1/2 O2 —> 4CO2 + 5H2O

4 0
2 years ago
A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t
jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x  (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is  initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for  weak acids for the equilibria:

protonated lidocaine + H₂O   ⇆  lidocaine + H₃O⁺

                      protonated lidocaine          lidocaine        H₃O⁺

Initial(M)               0.22                                       0                  0

Change                   -x                                      +x                 +x

Equilibrium          0.22 - x                                  x                    x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] =  x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸  = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22  - x  ≈ 0.22

and solve for x. The pH  will then  be the negative log of this value.

8.71 x 10⁻⁷  = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0
3 years ago
The solublity rules for ionic compounds are shown in the textbook on page 330. However, as you know, you can't bring the textboo
Alex Ar [27]

Answer:

NiS insoluble

Mg₃(PO₄)₂ insoluble

Li₂CO₃ soluble

NH₄Cl soluble

C₁₂H₂₂O₁₁ molecules

Explanation:

<em>Predict whether the following compounds are soluble or insoluble in water.</em>

Based on the solubility rules we can say:

  • NiS: Sulfides of transition metals are highly insoluble.
  • Mg₃(PO₄)₂: All phosphates (except those with metals of Group 1) are insoluble so Mg₃(PO₄)₂ is insoluble.
  • Li₂CO₃: all salts of metals of Group 1 are soluble so Li₂CO₃ is soluble.
  • NH₄Cl: all salts of ammonium are soluble so NH₄Cl is soluble.

<em>Which of the following best describes the solute in an aqueous solution of sucrose or C₁₂H₂₂O₁₁(aq)?</em>

Sucrose is a molecular compound in which atoms are linked through covalent bonds. Thus, it does not ionize in water (is a non-electrolyte) and when it dissolves it exists as C₁₂H₂₂O₁₁ molecules.

7 0
3 years ago
How many atoms are there in:
hoa [83]
B is the right answer "right"
3 0
2 years ago
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