Make the following conversion. 954 g= _ kg 0954000 95400 0.954 0.0954

Polymerization starts at the ________ end of the template.

balandron [24]

Answer:

B. 3'

Explanation:

Polymerization is the process of forming large molecules by joining together many small molecules. The small molecules are known as the monomers, and the macromolecules are called polymers.

There are two(2) types of polymerization.

a. addition polymerization

b. condensation polymerization

Polymerization is characterized by eliminating small molecules such as water, carbon(IV)oxide, methanol e.t.c.

In complex compounds, polymerization starts at the 3' end of the template.

3 0

3 years ago

How many grams (m) of glucose are in 165 ml of a 5.50% (m/v) glucose solution? express your answer numerically in grams?

Leya [2.2K]

mass of glucose = 0.055 *165 = 9.075 g

vol of methyl alc = 0.185 * 1.87 = 0.346 L = 346 ml

% NaCl ( m/v ) = mass NaCl * 100/ vol of soln

or Vol of Soln = mass NaCl / % NaCl (m/v)

= 32.1 * 100 / 6 = 535 ml the total vol of soln

6 0

3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

2 years ago

For the following electrochemical reaction: Al3+(aq) + 3e -> Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -> 2F (aq) Calculate

makvit [3.9K]

Answer: The standard electrode potential of the cell is 4.53 V.

Explanation:

We are given:

$E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=2.87-(-1.66)=4.53V$

Hence, the standard electrode potential of the cell is 4.53 V.

6 0

2 years ago

How is the mantle different from the core?

Pepsi [2]

Answer:The Earth's mantle is made up of semisolid rocks. ... The core consists of extremely hot metal layers instead of rock. Iron and nickel make up the outer section of the core, while the interior is almost entirely iron. The inner core is almost totally solid and shaped like a ball.

Explanation:

3 0

3 years ago