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mariarad [96]
3 years ago
9

Nahco3 is the active ingredient in baking soda. how many grams of oxygen are in 0.72 g of nahco3?

Chemistry
1 answer:
maksim [4K]3 years ago
4 0
M (O) = M x n

M (O) = 16g/mol

n (O) = 3 x n (NaHCO3)

n (NaHCO3) = m/M

m (NaHCO3) = 0.72g

M (NaHCO3) = 23 + 1 + 14 + 16 x 3 = 86g/mol

n (NaHCO3) = 0.72/86 = 0.008372mol

n (O) = 3 x 0.008372 = 0.02512

m (O) = 16 x 0.02512 = 0.402g
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Abbreviation for mole
Sedbober [7]

Answer:

Maybe mol

Explanation:

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Which statement best describes what is taking place?
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Therefore Chlorine is losing electrons and being oxidized. Hope it helps.

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Write out the balanced equation for the reaction that occurs when Al(OH)3 and H2SO4 react together. You do not need to make your
zhannawk [14.2K]
When Al(OH)3 and H2SO4  are allowed to react, the products formed are aluminum sulfate (Al2(SO4)3<span> ) and water (H2O). The balanced equation should be written as follows:

</span><span>3H2SO4 + 2Al(OH)3 = Al2(SO4)3 + 6H2O

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7 0
4 years ago
The density of an unknown gas at 98°C and 740 mmHg is 2.50 g/L. What is the molar mass of the gas with work showed?
noname [10]

Answer:

78.2 g/mol  

Step-by-step explanation:

We can use the <em>Ideal Gas Law</em> to solve this problem:

       pV = nRT

Since n = m/M, the equation becomes

      pV = (m/M)RT     Multiply each side by M

   pVM = mRT               Divide each side by pV

        M = (mRT)/(pV)

Data:

ρ = 2.50 g/L

R = 0.082 16 L·atm·K⁻¹mol⁻¹

T =98 °C

p = 740 mmHg

Calculation:

(a)<em> Convert temperature to kelvins </em>

T = (98 + 273.15) = 371.15 K

(b) <em>Convert pressure to atmospheres </em>

p = 740 × 1/760 =0.9737 atm

(c) <em>Calculate the molar mass </em>

Assume V = 1 L.

   Then m = 2.50 g

            M = (2.50 × 0.082 06 × 371.15)/(0.9737 × 1)

                = 76.14/0.9737

                = 78.2 g/mol

3 0
3 years ago
Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 10.0 kJ or greater.
nikdorinn [45]

Answer:

The answer to this can be arrived at by clculating the mole fraction of atoms higher than the activation energy of 10.0 kJ by pluging in the values given into the Arrhenius equation. The answer to this is 20.22 moles of Argon have energy equal to or greater than 10.0 kJ

Explanation:

From Arrhenius equation showing the temperature dependence of reaction rates.

K = Ae^{\frac{Ea}{RT} } where

k = rate constant

A = Frequency or pre-exponential factor

Ea   =       energy of activation

R = The universal gas constant

T = Kelvin absolute temperature

we have

f = e^{\frac{Ea}{RT} }

Where

f = fraction of collision with energy higher than the activation energy

Ea = activation energy = 10.0kJ = 10000J

R = universal gas constant = 8.31 J/mol.K

T = Absolute temperature in Kelvin = 400K

In the Arrhenius equation k = Ae^(-Ea/RT), the factor A is the frequency factor and the component e^(-Ea/RT) is the portion of possible collisions with high enough energy for a reaction to occur at the a specified temperature  

Plugging in the values into the equation relating f to activation energy we get

f = e^{\frac{10000J}{(8.31J/((mol)(K)))(400K)} } or f = e^{3.01} = 20.22 moles of argon have an energy of 10.0 kJ or greater

5 0
3 years ago
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