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EastWind [94]
4 years ago
12

You will analyze four substances in this lab. Based on their chemical formulas and what you already know about covalent and ioni

c compounds, make a prediction for each compound.Cornstarch, a carbohydrate consisting of hydrogen, carbon, and oxygen
Chemistry
1 answer:
In-s [12.5K]4 years ago
4 0

Answer:

Oil: covalent

Cornstarch: Covalent

Sodium chloride: Ionic

Sodium bicarbonate: Ionic

Explanation:

Covalent compounds are formed by sharing of electrons between non metals whereas ionic compounds are formed by transfer of electrons from metals to non metals.

1. Oil, which is built from the nonmetals hydrogen, carbon, and oxygen: forms a covalent compound by sharing of electrons between non metals hydrogen, carbon, and oxygen. Covalent compounds are insoluble in water.

2. Cornstarch, a carbohydrate consisting of hydrogen, carbon, and oxygen: forms a covalent compound by sharing of electrons between non metals hydrogen, carbon, and oxygen. Covalent compounds are insoluble in water.

3. Sodium chloride (table salt), whose formula is NaCl is formed by transfer of electrons from sodium to chlorine.Ionic compounds are soluble in water.

4. Sodium bicarbonate, whose formula is is formed by transfer of electrons from sodium to .Ionic compounds are soluble in water.

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PLZ HELP ...The half-life of polonium-218 is 3.0 minutes. If you start with 40.0 g, how long will it be before only 4.5 g remain
Elanso [62]

Answer:-  9.4 minutes.

Solution:- Radioactive decay obeys first order reaction kinetics and the equation used to solve this type of problems is:

lnN=-kt+lnN_0

where, k is decay constant and t is the time. N_0 is the initial amount of the radioactive substance and N is the remaining amount.

Since the value of decay constant is not given, so we need to calculate it first from given half life by using the formula:

k=\frac{0.693}{t_1_/_2}

where t_1_/_2 stands for half life.

Given half life is 3.0 minutes.

So, k=\frac{0.693}{3.0min}

k=0.231min^-^1

Let's plug in the values in the first order reaction equation and solve it for t.

ln4.5g=-0.231min^-^1(t)+ln40.0g

It could also be written as:

ln(\frac{4.5g}{40.0g})=-0.231min^-^1}

-2.18=-0.231min^-^1}

t=\frac{-2.18}{-0.231min^-^1}

k = 9.4 min

So, the radioactive substance would take 9.4 minutes to decay from 40.0 grams to 4.5 grams.

4 0
3 years ago
Biological membranes are present in all cells, and they make up the endomembrane system of eukaryotic cells. Among other functio
docker41 [41]

Answer:

The correct option for question 1 would be:

a bilayer containing lipids with hydrophilic head groups pointing inward and hydrophobic tail groups facing the solvent (extracellular fluid and cytosol).

The correct option for question number two would be: proteins.

Explanation:

The membranes present phospholipids that act as selective barriers between the intracellular and extracellular space, allowing an internal balance in relation to the external one.

Its conformation is mostly phospholipids, fatty acids, proteins (from transmembrane to intermembrane or external)

5 0
3 years ago
Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers known. It catalyzes the reversible hydrat
Afina-wow [57]

Explanation:

According to the given data, the turnover number can be calculated as follows.

      Turnover number = K_{cat} = \frac{V_{max}}{\text{Concentration of enzyme}}

     V_{max} = \frac{\text{Moles of CO_{2} hydrolyzed}{second}

Therefore, moles of CO_{2} hydrolyzed is as follows.

Moles of CO_{2} hydrolyzed = \frac{Mass of CO_{2}}{Molar mass of CO_{2}}

                 = \frac{0.30}{44}

                 = 0.00682 moles

Now, moles of CO_{2} hydolyzed per second is calculated as follows.

Moles of CO_{2} hydolyzed per second = \frac{0.00682}{60}

             = 1.137 \times 10^{-4} moles/second = V_{max}

And,

Moles of enzyme = \frac{Mass}{\text{Molar mass}}

                       = \frac{10.0 \mu g}{30000}

                       = 3.33 \times 10^{-10} moles

Therefore, the value of K_{cat} is as follows.

    K_{cat} = \frac{1.137 \times 10^{-4} moles}{3.333 \times 10^{-10} moles}

               = 0.3411 \times 10^{6} per second

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               = 20.466 \times 10^{6} per minute

Thus, we can conclude that the turnover number (K_{cat}) of carbonic anhydrase (in units of min^{-1}) is 20.466 \times 10^{6} per minute.

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