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AnnZ [28]
3 years ago
5

Which statement compares the masses of two subatomic particles?

Chemistry
1 answer:
Nataly [62]3 years ago
6 0
<span>The mass of a proton is greater than the mass of an electron.</span>
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Gaseous carbon monoxide reacts with hydrogen gas to form gaseous methane (ch4) and liquid water. express your answer as a chemic
laiz [17]
The formula for the compounds in the reaction are as follows with the respective states 
Carbon monoxide - CO (g)
hydrogen - H₂ (g)
methane - CH₄(g)
water - H₂O (l)
reaction of carbon monoxide with hydrogen gas gives rise to methane and water
the balanced chemical equation for the above reaction is as follows
CO(g) + 3H₂(g)  --> CH₄(g) + H₂O(l)
8 0
3 years ago
Positron emission of technetium-95
NNADVOKAT [17]
Idk it’s making me answer a question in order to get help on my own
5 0
3 years ago
f 23.6 mL of 0.200 M NaOH is required to neutralize 10.00 mL of a H3PO4 solution , what is the concentration of the phosphoric a
Eduardwww [97]

Answer:

Explanation:

H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)

mole of NaOH = 23.6 * 10 ⁻³L * 0.2M

= 0.00472mole

let x be the no of mole of H3PO4 required of  0.00472mole of NaOH

3 mole of NaOH required ------- 1 mole of H3PO4

0.00472mole of NaOH ----------x

cross multiply

3x = 0.0472

x = 0.00157mole

[H3PO4] = mole of H3PO4 / Vol. of H3PO4

= 0.00157mole / (10*10⁻³l)

= 0.157M

<h3>The concentration of unknown phosphoric acid is  0.157M</h3>
7 0
2 years ago
Read 2 more answers
An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an
Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that pH = -log[H^{+}], and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

pH = pKa - log \frac{[HA]}{[A^{-}]}

Where [HA] is the concentration of the acid, and [A^{-}] is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same  stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of OH^- will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of OH^- reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of OH^-, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0
3 years ago
The ksp of yttrium fluoride, yf3, is 8.62 × 10-21. calculate the molar solubility of this compound.
motikmotik

Answer:

The molar solubility of YF₃ is 4.23 × 10⁻⁶ M.

Explanation:

In order to calculate the molar solubility of YF₃ we will use an ICE chart. We identify 3 stages: Initial, Change and Equilibrium and we complete each row with the concentration of change of concentration. Let's consider the solubilization of YF₃.

       YF₃(s) ⇄ Y³⁺(aq) + 3 F⁻(aq)

I                       0               0

C                     +S            +3S

E                       S              3S

The solubility product (Ksp) is:

Ksp = [Y³⁺].[F⁻]³= S . (3S)³ = 27 S⁴

S=\sqrt[4]{Ksp/27} =\sqrt[4]{8.62 \times 10^{-21}  /27}=4.23 \times 10^{-6}M

6 0
3 years ago
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