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3 years ago

How many moles are in 110 grams of nahco3

1 answer:

4 0

Molar mass NaHCO₃ = 23 + 1 + 12 + 16 x 3 = 84 g/mol

1 mole ---------- 84 g
? mole ---------- 110 g

moles NaHCO₃ = 110 . 1 / 84

moles NaHCO₃ = 110 / 84

= 1.309 moles

hope this helps!

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Answer: Sodium perchlorate

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3 years ago

¿Se puede disolver acido sulfurico en NaCl ?

lord [1]

Answer:

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3 years ago

the reaction a(g) ⇌ b(g) has an equilibrium constant that is less than one. what can you conclude about ∆g° for the reaction?

-Dominant- [34]

For this reaction: ΔG⁰>0.

Balanced chemical reaction A(g) ⇌ (g)

ΔG° indicates that all reactants and products are in their standard states.

ΔG° = R·T·lnK.

ΔG° is Gibbs free energy

T is the temperature on the Kelvin scale

R is the ideal gas constant

The equilibrium constant (K) is the ratio of the partial pressures or the concentrations of products to reactants.

Gibbs free energy (G) determines if reaction will proceed spontaneously, nonspontaneously or in equilibrium processes.

If K < 1, than ΔG° > 0.

Reactants (in this example A) are favored over products (in this example B) at equilibrium.

More about equilibrium: brainly.com/question/25651917:

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4 0

1 year ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

An atom has the following electron configuration. 1s22s22p63s23p64s? How many valence electrons does this atom have? A: 2B: 3C:

Katena32 [7]

ANSWER

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STEP-BY-STEP EXPLANATION:

The electronic configuration is given below as

$1s^22s^22p^62s^23p^64s^2$

Looking at the configuration, you will see that the last energy carries 2 electrons

Hence, the valence electrons are 2

3 0

1 year ago