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3 years ago

Question 24

Chemistry

2 answers:

artcher [175]3 years ago

7 0

Answer: false
——————-

Brut [27]3 years ago

5 0

False, the energy from your food turns into glucose

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Radioactive decay can be described by the following equation where is the original amount of the substance, is the amount of the

soldi70 [24.7K]

Answer:

Iron remains = 17.49 mg

Explanation:

Half life of iron -55 = 2.737 years (Source)

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{2.737}\ year^{-1}$

The rate constant, k = 0.2533 year⁻¹

Time = 2.41 years

$[A_0]$ = 32.2 mg

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$[A_t]=32.2\times e^{-0.2533\times 2.41}\ mg$

$[A_t]=32.2\times e^{-0.610453}\ mg$

$[A_t]=17.49\ mg$

<u>Iron remains = 17.49 mg</u>

8 0

3 years ago

The two naturally occuring isotopes of antimony are 121Sb (57.21%) and 123Sb (42.79%), with isotopic masses of 120.904 and 122.9

emmasim [6.3K]

Answer:

The average atomic weight = 121.7598 amu

Explanation:

The average atomic weight of natural occurring antimony can be calculated as follows :

To calculate the average atomic mass the percentage abundance must be converted to decimal.

121 Sb has a percentage abundance of 57.21%, the decimal format will be

57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .

123 Sb has a percentage abundance of 42.79%, the decimal format will be

42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .

Next step is multiplying the fractional abundance to it masses

121 Sb = 0.5721 × 120.904 = 69.169178400

123 Sb = 0.4279 × 122.904 = 52.590621600

The final step is adding the value to get the average atomic weight.

69.169178400 + 52.590621600 = 121.7598 amu

5 0

3 years ago

Complete combustion of 5.90 g of a hydrocarbon produced 19.2 g of CO2 and 5.89 g of H2O. What is the empirical formula for the h

scZoUnD [109]

$44g \ CO_{2} \ \ \ \rightarrow \ \ 12g \ C\\
19,2g \ CO_{2} \rightarrow \ \ \ x\\\\
x=\frac{19,2g*12g}{44g}\approx5,24g \ \ \ \Rightarrow \ \ n=\frac{5,24g}{12\frac{g}{mol}}\approx0,44mol\\\\\\
18g \ H_{2}O \ \ \ \rightarrow \ \ \ 2g \ H\\
5,89g \ H_{2}O \ \rightarrow \ \ y\\\\
y=\frac{5,89g*2g}{18g}\approx0,65g \ \ \ \ \Rightarrow \ \ n=\frac{0,65g}{1\frac{g}{mol}}=0,65mol\\\\\\
n_{C}:n_{H}=0,44:0,65\approx1:1\\\\
CH$

4 0

3 years ago

Choose the phrase that best defines ecosystem service. (1 point)

Fantom [35]

Human:

✧･ﾟ: *✧･ﾟ:* Answer: *:･ﾟ✧*:･ﾟ✧

✅ the first one

I WOULD APPRECIATE BRAINLIEST!

~ ₕₒₚₑ ₜₕᵢₛ ₕₑₗₚₛ! :₎ ♡

7 0

3 years ago

Read 2 more answers

Describe how water molecules interact with sodium chloride molecules in order for water to dissolve salt.

Phantasy [73]

Answer:

Sodium chloride is an ionic salt that completely dissociates in water.

Explanation:

Sodium chloride in water dissociates according to the equation.

NaCl (aq) → Na + (aq) + Cl- (aq)

While water stays in balance according to the equation

2H2O (l) ⇄ H3O+(aq) + OH-(aq).

By dissociating the solute in the solvent, the water molecules pick up the sodium cations as well as the chloride anions to break the bonds in the salt crystals and thus produce the dissolution

If the ions are analyzed separately, we see that Na + interacts with water, so it would be producing sodium hydroxide, releasing protons.

If the reaction goes in the sense of products, the Na + would increase the concentration of [H +], however being a neutral salt the pH does not vary from 7, so the reaction would never go in the sense of products, but only reagents. That's why as the sodium hydroxide is a strong base, it only promotes the formation of water and the cation.

Na+ + H2O ← NaOH + H+

The same situation occurs with chloride, since no matter how much you want to form hydrochloric acid, it dissociates by promoting the formation of water plus the corresponding anion. Na + and Cl- act as too weak conjugate acid and base, so the reaction would never go in the direction of products

Cl- + H2O ← HCl + OH-

3 0

4 years ago