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2 years ago

Ca(hco3)2 molecular formula calculated

Chemistry

1 answer:

GrogVix [38]2 years ago

7 0

Answer:

C2H2CaO6

Explanation:

Ca same

H × 2

C × 2

O3 × 2

I hope this helps you :)

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2 years ago

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A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

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2 years ago

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The concentration of protein in a urine sample is calculated to be 2.77 μg/mL. What is the concentration of this solution in uni

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Answer:

The concentration of this solution in units of pounds per gallon is $2.776*10^{-5} \frac{lb}{gal}$

Explanation:

Units of measurement are established models for measuring different quantities. The conversion of units is the transformation of a quantity, expressed in a certain unit of measure, into an equivalent one, which may or may not be of the same system of units.

In this case, the conversion of units is carried out knowing that 1 μg are equal to 2.205*10⁻⁹ Lb and 1 mL equals 0.00022 Gallons. So

$2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}$

If 1 μg equals 2.205*10⁻⁹ lb, 2.77 μg how many lb equals?

$lb=\frac{2.77ug*2.205*10^{-9}lb }{1ug}$

lb=6.10785*10⁻⁹

So, 2.77 μg= 6.10785*10⁻⁹ lb

Then:

$2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}=\frac{6.10785*10^{-9}lb }{mL} =\frac{6.10785*10^{-9}lb }{0.00022 gal} =\frac{6.10785*10^{-9}lb }{0.00022 gal}$

You get:

$2.77 \frac{ug}{mL} = 2.776*10^{-5} \frac{lb}{gal}$

The concentration of this solution in units of pounds per gallon is $2.776*10^{-5} \frac{lb}{gal}$

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2 years ago

8.) If 396 g of Carbon Dioxide (CO2) are produced, what mass of Oxygen (02) reacted? *

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Answer:

264g

Explanation:

C + O2 -> CO2

_g + _g -> 396g

396÷3=132

C (132g) + O2 (264g) -> CO2 (396g)

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2 years ago

Calculate the number of grams of nitrogen dioxide that are produced from 4 moles of nitric oxide

miv72 [106K]

Answer:

hope this help by the way found off of yahoo

Explanation:

Calculate the number of grams of nitrogen dioxide that are produced from

4 moles of nitric oxide.

2NO(g) + O2(g) -->2NO2(g)

I really need help with this... I need to know how to work it too... I can balance it out but not sure about grams... This is it balanced out with 4 moles of nitric oxide

4NO(g) + 2O2(g) ->4NO2(g) please help and explain i want to learn this

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2 years ago

Ca(hco3)2 molecular formula calculated​

Ca(hco3)2 molecular formula calculated