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kotykmax [81]
2 years ago
13

Express the following numbers in Standard Form:

Chemistry
1 answer:
guajiro [1.7K]2 years ago
8 0

Answer:

0.0027

28,100,000

0.0000367894

456,760

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The ratio would be 12.6 of carbon acid
7 0
3 years ago
chemistry A basketball is inflated to a pressure of 1.10 atm in a 28.0°C garage. What is the pressure of the basketball outside
butalik [34]

Answer : The final pressure of the basketball is, 0.990 atm

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

or,

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

where,

P_1 = initial pressure = 1.10 atm

P_2 = final pressure = ?

T_1 = initial temperature = 28.0^oC=273+28.0=301.0K

T_2 = initial temperature = -2.00^oC=273+(-2.00)=271.0K

Now put all the given values in the above equation, we get:

\frac{1.10atm}{P_2}=\frac{301.0K}{271.0K}

P_2=0.990atm

Thus, the final pressure of the basketball is, 0.990 atm

7 0
2 years ago
2 attempts left Click the "draw structure" button to launch the drawing utility. Draw one product formed when the following dien
Yuliya22 [10]

Answer:

2-chloro-1-methyl-cyclohex-1,4-diene.

Explanation:

Hello,

In this case, the addition of hydrochloric acid acts as an electrophilic atack in which the hydrogen bonded to the double-bonded carbon connected to the carbon with the methyl substitution is substituted by the chlorine from the hydrochloric acid, in such a way, 2-chloro-1-methyl-cyclohex-1,4-diene is produced as one equivalent of HCl is used therefore one substitution will be attained for chlorine, and hydrogen as a side product as shown on the attached picture.

Best regards.

7 0
3 years ago
Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration
elixir [45]

Answer:

8.54

Explanation:

At equivalence point :  

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052             = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa  

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :  

pH = 1/2 [ pKw + pKa + log c ]  

Ka for acetic acid = 1.75 X 10⁻⁵  

so pKa = -log (1.75 X 10⁻⁵) = 4.74  

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169  

so log c = log 0.02169 = -1.66  

putting the values....  

pH = 1/2 [14 + 4.74 - 1.66 ]  

pH = 1/2 [ 17.08] = 8.54

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