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Fynjy0 [20]
3 years ago
10

Determine the mass of 2.75 moles of CaSO4. Record your work and your answer.

Chemistry
1 answer:
maw [93]3 years ago
6 0

Explanation:

RFM \:  = 160 \: g \\ 1 \: mole \: weighs \: 160 \: g \\ 2.75 \: moles \: weighs \:  \frac{2.75 \times 160}{1} g \\  = 440 \: g

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The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O represents an acid-base tritration. The equivalence point occurs when 24.75 mL o
artcher [175]
            moles NaOH = c · V = 0.2432 mmol/mL · 24.75 mL = 6.0192 mmol
            moles H2SO4 = 6.0192 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.0096 mmol
Hence
            [H2SO4]= n/V = 3.0096 mmol / 38.94 mL = 0.07729 M
The answer to this question is  [H2SO4] = 0.07729 M

6 0
3 years ago
Scientists frequently use all of the following to support the theory that organisms change over time except.
Pie

Answer:

4

Explanation:

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3 years ago
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Scientists believe that there are three genes that contribute to skin color in humans.
Darina [25.2K]

Answer:B

Explanation:

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3 years ago
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A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a
Mandarinka [93]
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA] 

 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid

Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol

Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles

pH = 4.74 + log (.115/0.085)
pH = 4.87
6 0
3 years ago
Calculate the pH in titration of a weak acid: What is the pH in titration of formic acid (HCHO2, 0.200 M, 100.0 mL) after the ad
ki77a [65]

Answer:

pH = 12.61

Explanation:

First of all, we determine, the milimoles of base:

0.120 M = mmoles / 300 mL

mmoles = 300 mL . 0120 M = 36 mmoles

Now, we determine the milimoles of acid:

0.200 M = mmoles / 100 mL

mmoles = 100 mL . 0.200M = 20 mmoles

This is the neutralization:

HCOOH    +     OH⁻         ⇄        HCOO⁻     +    H₂O

20 mmol       36 mmol             20 mmol

                    16 mmol

We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:

NaHCOO  →  Na⁺  +  HCOO⁻

HCOO⁻  +  H₂O  ⇄   HCOOH  +  OH⁻   Kb →  Kw / Ka = 5.55×10⁻¹¹

These contribution of OH⁻ to the solution is insignificant because the Kb is very small

So:  [OH⁻] =  16 mmol / 400 mL →  0.04 M

- log  [OH⁻]  = pOH →  1.39

pH = 14 - pOH → 12.61

6 0
3 years ago
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