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mote1985 [20]
3 years ago
13

Theoretically, how many moles of carbonic acid will be produced by 3.00 g sample of NaHCO3?

Chemistry
1 answer:
Kryger [21]3 years ago
7 0
NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol 

Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27% 

3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3 

0.8211 grams Na + 1.266 grams Cl = 2.087 grams
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zavuch27 [327]

Answer: M = 22/ (i x28.948)

Explanation:

Pi = osmotic pressure = 22atm

T = Temperature = 353K

M = Molarity = ?

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i = van’t Hoff factor

Pi = iMRT

M= Pi /(iRT) = 22 / ( i x 0.082 x 353)

M = 22/ (i x28.948)

7 0
3 years ago
Given the speed of light (3.00 × 108 m/s) and a wavelength 5.9 × 10-7 m, what is the frequency?
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Wavelength x Frequency = Speed

So, 5.9 x 10^-7 x F = 3 x 10^8

Rearranging the equation, we get:
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3 years ago
Which best explains why blood is considered to be a suspension?
Stells [14]
I think the best answer is b): it contains particles that can settle out.

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5 0
3 years ago
If a solution initially contains 0.260 M HC2H3O2, what is the equilibrium concentration of H3O+ at 25 ∘C? Express your answer in
arlik [135]

Answer:

2.16 × 10⁻³

Explanation:

Step 1: Given data

Concentration of the acid (Ca): 0.260 M

Acid dissociation constant (Ka): 1.80 × 10⁻⁵

Step 2: Write the acid dissociation equation

HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)

Step 3: Calculate the concentration of H₃O⁺ at equilibrium

We will use the following expression.

[H_3O^{+} ]= \sqrt{Ka \times Ca } = \sqrt{1.80 \times 10^{-5} \times 0.260 } = 2.16 \times 10^{-3}

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3 years ago
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Yuri [45]

Answer:

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