All categories

3 years ago

A compound is 40% carbon, 6.7% hydrogen, and 53.3% oxygen. What is the empirical formula?

Chemistry

2 answers:

Ahat [919]3 years ago

6 0

Answer:

O H C

Moles in 100g 3.33 6.65 3.33

Ratio 1.00 2.00 1.00

Possible empirical formula = $OH_{2}C$

Oduvanchick [21]3 years ago

4 0

Answer:

CH2O

Explanation:

Data obtained from the question include:

C = 40%

H = 6.7%

O = 53.3%

Divide by their molar mass

C = 40/12 = 3.33

H = 6.7/1 = 6.7

O = 53.3/16 = 3.33

Divide by the smallest

C = 3.33/3.33 = 1

H = 6.7/3.33 = 2

O = 3.33/3.33 = 1

The empirical formula is CH2O

You might be interested in

What is the unit used for water vapor?

SpyIntel [72]

Answer:

relative humidity

Explanation:

Relative humidity can be measured with a hygrometer

4 0

3 years ago

Which of the following is not a step in the conversion of 1.3 dam3 to L?

Tanya [424]

First put the choices,and the answer is <span>(10^-3L/1mL)^3 is the correct answer.</span>

3 0

3 years ago

EXPERIMENT 4 : TYPES OF PLATE BOUNDARY

zalisa [80]

Answer:

Movement in narrow zones along plate boundaries causes most earthquakes. Most seismic activity occurs at three types of plate boundaries—divergent, convergent, and transform. As the plates move past each other, they sometimes get caught and pressure builds up.

Explanation:

3 0

2 years ago

Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 oC) in the absence of a ca

Oksi-84 [34.3K]

Answer:

30 kJ

Explanation:

Arrhenius equation is given by:

$k=Aexp(-Ea/RT)\\$

Here, k is rate constant, A is Pre-exponential factor, Ea is activation energy and T is temperature.

taking natural log of both side

ln k = ln A - Ea/RT

In Arrhenius equation, A, R and T are constant.

Therefore,

$ln\frac{k_2}{k_1} =\frac{Ea_1-Ea_2}{RT}$

$Ea_1-Ea_2$ is the lowering in activation energy by enzyme,

R = 8.314 J/mol.K

T = 37°C + 273.15 = 310 K

$\frac{k_2}{k_1} =1\times 10^5$

$ln 1\times 10^5 =\frac{Ea_1-Ea_2}{RT}\\{Ea_1-Ea_2} = 11.512 \times 8.314 \times 310\\=29670\ J\\=30\ kJ$

4 0

3 years ago

A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 3.30kg of water at 23.8°C. Dur

Brums [2.3K]

Answer:

$T_2= 31.9\°C$

Explanation:

Hello there!

In this case, it is possible to propose an energy balance in order to illustrate how the heat released by the reaction is absorbed by the water:

$-Q_{rxn}=Q_{water}$

Thus, since the heat released by the reaction is -112 kJ (-112000 J), it is possible to define the hear absorbed by the water in terms of mass, specific heat and temperature change:

$-(-112000J)=m_{water}C_{water}(T_2-T_1)$

In such a way, it is possible to define the final temperature as shown below:

$T_2=23.8\°C+\frac{112000J}{3300g*4.18\frac{J}{g\°C} }\\\\T_2= 31.9\°C$

Best regards!

4 0

2 years ago