How is the half-life of a radioactive material useful for radioactive dating?

Explain why decomposition of sugar on heating is an irreversible change.

Elena L [17]

Sugar, sucrose (C12H22O11: a disaccharide, composed of the two monosaccharides: glucose and fructose), is odorless, that is, it lacks odor. When heated a phase change occurs resulting in melting of a thick syrup.

5 0

3 years ago

how many moles of carbon monoxide can be produced from 8 moles of carbon if carbon is the limiting reagent?

Mademuasel [1]

Answer:

8 moles of CO

Explanation:

To produce carbon monoxide we begin from C and O₂ as this reaction shows:

2C + O₂ → 2CO

Therefore, 2 moles of C can produce 2 moles of CO

If I have 8 moles of C, I must produce 8 moles of CO

Ratio is 1:1

6 0

3 years ago

You are given the reaction Cu + HNO3 Cu(NO3)2 + NO + H2O complete the final balanced equation based on half-reactions

Evgen [1.6K]

Answer:

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

Explanation:

Cu + HNO3 → Cu(NO3)2 + NO + H2O

The first step is to write the oxidation numbers for each atoms in the given equation

Cu0 + H+1N+5O-23 → Cu+2(N+5O-23)2 + N+2O-2 + H+12O

Identify the oxidizing and reducing agent

OXIDATION --- Cu0 → Cu+2(N+5O-23)2 + 2e-

REDUCTION---H+1N+5O-23 + 3e- → N+2O

Balance equation in half reaction

Cu0 + 2HNO3 → Cu+2(N+5O-23)2 + 2e-

H+1N+5O-23 + 3e- → N+2O

Now balance the charge

Cu0 + 2HNO3 → Cu+2(N+5O-23)2 + 2e- + 2H+

H+1N+5O-23 + 3e- + 3H+ → N+2O

Balance the oxygen atom

Cu0 + 2HNO3 → Cu+2(N+5O-23)2 + 2e- + 2H+

H+1N+5O-23 + 3e- + 3H+ → N+2O-2 + 2H2O

Make electron gain equivalent to electron lost.

3Cu0 + 6HNO3 → 3Cu+2(N+5O-23)2 + 6e- + 6H+

2H+1N+5O-23 + 6e- + 6H+ → 2N+2O-2 + 4H2O

Complete reaction

3Cu0 + 8H+1N+5O-23 + 6e- + 6H+ → 3Cu+2(N+5O-23)2 + 2N+2O-2 + 6e- + 4H2O + 6H+

Simplify the equation

3Cu0 + 8H+1N+5O-23 → 3Cu+2(N+5O-23)2 + 2N+2O-2 + 4H2O

Final equation

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

5 0

3 years ago

A chemistry graduate student is given 125.mL of a 0.20M acetic acid HCH3CO2 solution. Acetic acid is a weak acid with =Ka×1.810−

Over [174]

Answer : The mass of sodium acetate is, 1.097 grams.

Explanation : Given,

The dissociation constant for acetic acid = $K_a=1.8\times 10^{-5}$

Concentration of acetic acid (weak acid)= 0.20 M

volume of solution = 125. mL

pH = 4.47

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (1.8\times 10^{-5})$

$pK_a=5-\log (1.8)$

$pK_a=4.74$

Now we have to calculate the concentration of sodium acetate (conjugate base or salt).

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$4.47=4.74+\log (\frac{[Salt]}{0.20})$

$[Salt]=0.107M$

Now we have to calculate the mass of sodium acetate.

$\text{Concentration}=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{\text{Molar mass of }NaCH_3CO_2\times \text{Volume of solution (in mL)}}$

$0.107M=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{82g/mol\times 125mL}$

$\text{Mass of }NaCH_3CO_2=1.097g$

Therefore, the mass of sodium acetate is, 1.097 grams.

6 0

4 years ago

Batteries are a source of __________ energy.<br> chemical<br> potential<br> mechanical

dimaraw [331]

Answer:

Batteries store chemical energy

Explanation:

3 0

3 years ago

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