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PtichkaEL [24]
2 years ago
7

Write the molecular formula of poly(glycine)

Chemistry
1 answer:
Dvinal [7]2 years ago
5 0

Answer:

C2H5NO2

Explanation:

Glycine | C2H5NO2 - PubChem.

weight of glycine= 75.07 g/mol

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Question 044 Which of the following sequences converts 2-methylpropene and sodium acetylide into 3-methylbutanal? 1) HBr; 2) NaC
Minchanka [31]

Answer:

1) HBr; 2) NaCCH; 3) O3; 4) H2O

Explanation:

The first step is formation of alkyl halide followed by reaction with sodium acetylide, to form 3-methylbutene, which is then followed by oxidation reaction with O3& H2O to 3-methylbutanal

8 0
3 years ago
Read 2 more answers
The temperature is -14°C the air pressure in an automobile tire is 149K PA if the volume does not change what is the pressure af
Jet001 [13]

Answer: The pressure after the tire is heated to 17.3°C is 167 kPa

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=149kPa\\T_1=-14^0C=(273-14)=259K\\P_2=?=27.5psi\\T_2=17.3^0C=(273+17.3)=290.3K

Putting values in above equation, we get:

\frac{149}{259}=\frac{P_2}{290.3}\\\\P_2=167kPa

Hence, the pressure after the tire is heated to 17.3°C is 167 kPa

4 0
3 years ago
What is the IUPAC name
PolarNik [594]
Organic chemical compounds as recommended by the (IUPAC)
8 0
3 years ago
Does Kinetic energy depend only on an object’s speed and velocity?
BaLLatris [955]

yes         .. . . . .. . . .. ....... . ..dlnv3r;'mw,c    kc;oqc,

xkdnnd                                                                                                

4 0
3 years ago
Read 2 more answers
The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pres
Dima020 [189]

Answer:

6.25\times 10^{-6} is the value of the equilibrium constant at this temperature.

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures  of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as K_{p}

2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)

Partial pressures at equilibrium:

p^o_{H_2O}=0.070 atm

p^o_{H_2}=0.0035 atm

p^o_{O_2}=0.0025 atm

The equilibrium constant in terms of pressures is given as:

K_p=\frac{(p^o_{H_2})^2\times (p^o_{O_2})}{(p^o_{H_2O})62}

K_p=\frac{(0.0035 atm)^2\times 0.0025 atm}{(0.070 atm)^2}=6.25\times 10^{-6}

6.25\times 10^{-6} is the value of the equilibrium constant at this temperature.

5 0
3 years ago
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