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lianna [129]
3 years ago
14

When paper turns yellow-brown upon exposure to sunlight, what type of change is likely taking place?

Chemistry
1 answer:
sveticcg [70]3 years ago
8 0

the answer to this is chemical change

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A very old tree limb contains an amount of carbon-14 that is approximately 1/8 of the current atmospheric 14C levels.. Calculate
Tomtit [17]

A.       The radioactive decay equation is N = N0e^{-ln(2)*t/T }

where T is the half-life (5730 years), N0 is the number of atoms at time t = 0 and N is the number at time t.

Rewriting this as:

(N/N0) = e^{-ln(2)*t/T }

Since N = (1/8) N0 and substituting known values:

1/8 = e^{-ln(2)*t/5730}

Taking ln of both sides:

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3 0
3 years ago
When nahco3 completely decomposes, it can follow this balanced chemical equation: 2nahco3 → na2co3 h2co3 determine the theoretic
BigorU [14]

Theoretical yield = 2.397

The product could be sodium carbonate

percent yield = 98.456%

When nahco3 completely decomposes, it can follow this balanced chemical equation:

2nahco3 → na2co3 h2co3

If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.

mass of NaHCO₃ = 3.80 g

molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.80/84 =  0.0452 mol

You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.

so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol

∴ mass of sodium carbonate ( Na₂CO₃) = no of moles of Na₂CO₃ × molar mass of Na₂CO₃

=  0.0226 × 106 ≈ 2.397 g

no of moles of hydrogen carbonate = 0.0452/2 = 0.0226 mol

mass of the hydrogen carbonate ( H₂CO₃) = no of moles of H₂CO₃ × molar mass of H₂CO₃

= 0.0226 × 62 g = 1.401 g

mass of one of the products was measured to be 2.36 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.397 g.

percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.36 g

and theoretical yield of Na₂CO₃ = 2.397 g

∴ % yield = 2.36/2.397 × 100 ≈ 98.456%

Therefore the percentage yield of the product is 98.456%

To learn more about percentage yield visit:

brainly.com/question/22257659

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stronger than aluminum

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