When is a hypothesis useful in science?

A molecule has the empirical formula C4H6O. If its molecular weight is determined to be about 212 g/mol, what is the most likely

krok68 [10]

Answer:

The molecular formula is C12H18O3

Explanation:

Step 1: Data given

The empirical formula is C4H6O

Molecular weight is 212 g/mol

atomic mass of C = 12 g/mol

atomic mass of H = 1 g/mol

atomic mass of O = 16 g/mol

Step 2: Calculate the molar mass of the empirical formula

Molar mass = 4* 12 + 6*1 +16

Molar mass = 70 g/mol

Step 3: Calculate the molecular formula

We have to multiply the empirical formula by n

n = the molecular weight of the empirical formula / the molecular weight of the molecular formula

n = 70 /212 ≈ 3

We have to multiply the empirical formula by 3

3*(C4H6O- = C12H18O3

The molecular formula is C12H18O3

3 0

3 years ago

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What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)

blsea [12.9K]

<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, $(C_{2}H_{5})_{3}N$ , the reaction will be
$(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}$

and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value = 5.3 x 10^-4

And $kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}$

thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

=0.00516 M

Thus, pOH = 2.30
pH = 14 - pOH = 11.7

6 0

3 years ago

The gravitational force exerted by an object is given by F = mg, where F is the force in newtons, m is the mass in kilograms, an

Andre45 [30]

The height of the column is 0.457 m and the mass of the atmosphere is calculated as 1.03 × 10 ⁴ kg.

<h3>What is Pascal? </h3>

Pascal is defined as the force per unit area. It expressed in Newton pr square meter of area.

1 Pa = 1 N / m²

Pressure = force / Area

According to the question, the expression of force is as given below

F = mg

where,

F is the force,

m is the mass,

g is equal to the acceleration due to the gravity

Now, the area of the atmosphere is 1 m².The pressure is 1 atm. Pressure in Pascal.

1 atm = 1.01325 × 10 5pa

Therefore, the expression for pascal become as follows.

1.01325 × 10 5 pa = mg /area

1.01325 × 10 5 pa = m × 9.81 m/s² / 1 m²

M = 1.01325 × 10 5 pa × 1 m² / 9.81 m² × 1 Nm -² /1 pa × 1 kg m-² / 1 N

1.03 × 10 ⁴ kg

Given,

The density is 22.6 g /mL , pressure is 1 atm, and area is 1 m²

The relation between density and pressure can be given as follows.

P = hpg… … …(1 )

were , h is the height of the column

p is the density.

Hpg = 1.01325 × 10 5 pa × 1 N/m² /1 pa

H = 1.01325 × 10 5 N/m² / pg × 1 kg ms-² / 1 N

= 1.01325 × 10 5 kg m-¹ s -² / 22.6 g mL -1 × 1 kg/ 10 ³ g × 1 mL / 10 -6 m ³ × 9.81 m s- ²

= 0.457 m

Therefore, the height of the column is 0.457 m.

Thus, we concluded that the height of the column is 0.457 m and the mass of the atmosphere is calculated as 1.03 × 10 ⁴ kg.

learn more about density:

brainly.com/question/952755

#SPJ4

7 0

1 year ago

During an experiment, a scientist places a heat lamp above a bowl of water and uses the lamp to heat up the water. How does heat

dlinn [17]

Answer:

Radiation

Explanation:

6 0

2 years ago

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Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.

$pH=-log[H^{+}]$

So, $[H^{+}]=10^{-pH}$

Plug in all the given values in the equation of $Q_{p}$ :

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$

7 0

3 years ago