What is the purpose of finding oxidation states in the half-reaction method for balancing equations?

What is diabetes? How does it disrupt normal cellular activities? What are the overall consequences for your body?

Vlad1618 [11]

Answer:

Diabetes is a group of diseases that affect how your body uses blood sugar (glucose). Glucose is vital to your health because it's an important source of energy for the cells. Cells are everywhere including the blood, bones and muscles. When you have diabetes, you don’t have enough insulin in your blood (an organ in your body called the pancreas makes insulin). Insulin controls the movement of blood sugar into the cells of the body.

Explanation:

8 0

3 years ago

Determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.

AnnZ [28]

Answer: The limiting reagent is oxygen gas.

Explanation:

Limiting reagent is defined as the reactant that is present in less amount and it limits the formation of products.

Excess reagent is defined as the reactant which is present in large amount.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For propane:

Given mass of propane = 30.0 g

Molar mass of propane = 44.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of propane}=\frac{30.0g}{44.1g/mol}=0.680mol$

For oxygen:

Given mass of oxygen = 75.0 g

Molar mass of oxygen = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of propane}=\frac{75.0g}{32g/mol}=2.34mol$

The chemical equation for the combustion of propane follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$

By Stoichiometry of the reaction:

5 moles of oxygen gas reacts with 1 mole of propane.

So, 2.34 moles of oxygen gas will react with = $\frac{1}{5}\times 2.34=0.468mol$ of propane

As, given amount of propane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reagent is oxygen gas.

3 0

3 years ago

4.14 Oxygen requirement for growth on glycerol Klebsiella aerogenes is produced from glycerol in aerobic culture with ammonia as

Iteru [2.4K]

Answer:

0.37 g

Explanation:

The molecular weight for Glycerol = 92

Number of Carbon atoms in glycerol (x) $C_3H_8O_3$ = 3

Molecular weight of the Biomass ( Klebsiella aerogenes )

= $CH_{1.73}O_{0.43}N_{0.24}$

= $\frac{23.97}{0.92}$

= 26.1

From the molecular weight of the Biomass, we can deduce the Degree of reduction for the substrate(glycerol denoted as $\delta _g$ ) as follows:

= (4×1)+(1×1.73)-(2×0.43)-(3×0.24)

= 4.15

Given that the yield of the Biomass = 0.40 g

However;

C = $Yield of Biomass *\frac{Molecular weight of substrate}{Molecular weight of the Biomass}$

C = $0.40*\frac{92}{26.1}$

C = 1.41 g

Now , the oxygen requirement can be calculated as:

= $\frac{1}{4}*(n*S - C * \delta _{g})$

= $\frac{1}{4}(3*4.7-1.41*4.15)$

= 2.1 g/mol

Hence, we can say that the needed oxygen = 2.1 g/mol of the substrate consumed.

Now converting it to mass terms; we have:

= $2.1*\frac{number of mole of oxygen}{molecular weight of glycerol}$

= $2.1 * \frac{16}{92}$

= 0.3652 g

≅ 0.37 g

∴ The oxygen requirement for this culture in mass terms = 0.37 g

3 0

3 years ago

Balancing Practice (some of these may be in the simulation)

agasfer [191]

Answer:

see below

Explanation:

16. 1 P₄(s) + 6 F₂(g) → 1 PF₃(s)

17. 2 C(s) + 2 H₂O(g) → 1 CH₄(g) + 1 CO₂(g)

18. 2 HgO(s)→ 1 O₂(g) + 2 Hg(l)

19. 1 CaCO₃(s) → 1 CO₂(g) + 1 CaO(s)

8 0

3 years ago

What scientists do that is the basis for their investigation

Vlada [557]

They utilize and make use of the scientific method in order to have clear basis and evidence for their investigations. Research method is always used to answer every scientific inquiry and in gaining evidential data or knowledge. The scientific method has the following process or at least undergoes the process of
1. Observation
2. Hypothesis 
3. Experimentation 
4. Interpretation of data 
5. Evaluating the data 
6. Passing and recording the data 

These steps are crucial and the empirical data that these scientists obtain are very important to keep that is why research paper, thesis and dissertations exists.

5 0

4 years ago