All categories

3 years ago

What is chemistry and biochemistry

Chemistry

1 answer:

kow [346]3 years ago

3 0

Answer:

Chemistry mainly deals with study of physical and chemical properties of organic and inorganic matters but biochemistry involves mainly study of compounds (mostly organic compounds) of Biomedical interest.

You might be interested in

What is the major difference between chemical and physical changes?

viktelen [127]

C. New substances are always produced in chemical changes, which is not the case with physical changes. For example, when you cut wood into smaller pieces you still have wood in the end. But, with a chemical change you change the substance's whole composition. For example when you add two hydrogen's to one oxygen you have water and it can't break down from that.

5 0

3 years ago

A sample of water is mixed with a surfactant.what will most likely happen to the viscosity of the water?

GenaCL600 [577]

The viscosity will decrease

5 0

3 years ago

If this compounds (N2O) molecular mass is 88g are the empirical and molecular formulas the same as one another or different

Softa [21]

Answer:

Explanation:

Let us examine this problem carefully:

Given compound is N₂O

Molecular mass = 88g

Now,

The empirical formula is the simplest formula of a compound.

The molecular formula is the true formula of the compound that shows that actual ratios of the atoms in a compound.

To find if they both have the same molecular and empirical formula, they must have the same molecular mass.

For N₂O;

Molecular mass = 2(14) + 16 = 44g/mole

But the true and given molecular formula of the compound is 88g/mole

This shows that the compound given is the empirical formula of the compound.

Molecular formula:

molecular mass of empirical formula x n = molecular mass of molecular formula

n = $\frac{88}{44}$ = 2

Molecular formula of compound = 2(N₂O) = N₄O₂

Therefore the empirical and molecular formulas are not the same

8 0

4 years ago

If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?

ICE Princess25 [194]

Answer: The half life of the sample of silver-112 is 3.303 hours.

Explanation:

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

$k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}$

where,

k = rate constant = ?

t = time taken = 1.52 hrs

$[A_o]$ = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

$k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}$

To calculate the half life period of first order reaction, we use the equation:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life period of first order reaction = ?

k = rate constant = $0.2098hr^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs$

Hence, the half life of the sample of silver-112 is 3.303 hours.

6 0

3 years ago

lord [1]

Answer:

$\boxed{6.08 \times 10^{23}}$

Explanation:

Data:

I = 2.15 A

t = 8 min 24 s

T = 26.0 °C

V = 65.4 mL

p = 774.2 To

1. Write the equation for the half-reaction

2H₂O ⟶ O₂ + 4H⁺ + 4e⁻

2. Calculate the moles of oxygen

$p = \text{774.2 To} \times \dfrac{\text{1 atm}}{\text{760 To}} = \text{1.0187 atm}$

V = 0.0654 L

T = (26.0 + 273.15) K = 299.15 K

$\begin{array}{rcl}pV& = & nRT\\1.0189 \times 0.0654 & = & n \times 0.08206 \times 299.15\\0.06662 & = & 24.55n\\\\n & = & \dfrac{0.06662}{24.55}\\\\n & = & 2.714 \times 10^{-3}\\\end{array}$

3. Calculate the moles of electrons

$\text{n} = \text{2.714 $\times 10^{-3}$ mol oxygen} \times \dfrac{\text{4 mol electrons}}{\text{ 1 mol ozygen}} = \text{ 0.01086 mol electrons}$

4. Calculate the number of coulombs

t = 8 min 24 s =504 s

Q = It = 504 s × 2.10 C·s⁻¹= 1058 C

5. Calculate the number of electrons

$\text{No. of electrons} = \text{1058 C} \times \dfrac{\text{1 electron}}{1.602 \times 10^{-19}\text{ C}} = 6.607 \times 10^{21}\text{ electrons}$

6. Calculate Avogadro's number

$N_{\text{A}} = \dfrac{6.607 \times 10^{21}}{0.01086} = \mathbf{6.08 \times 10^{23}}\\\\\text{The experimental value of Avogadro's number is } \boxed{\mathbf{6.08 \times 10^{23}}}$

3 0

4 years ago

What is chemistry and biochemistry​

What is chemistry and biochemistry