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2 years ago

How many hydrogen atoms are represented by (CH3)2CHOH?

Chemistry

1 answer:

lozanna [386]2 years ago

3 0

Answer:

8 Hydrogen Atoms

Explanation:

The subscripts that are next to the coefficient (letter) indicates how much atoms the element/compound has.

So:

(CH3) × 2 ⇒ 6 hydrogen atoms

+ 1 from (CH) and another from (OH) ⇒ 8 atoms

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Convert the density of water, 1.00 g/mL to lb/ft3. Provide only the numerical value in your answer with the correct number of si

PtichkaEL [24]

Answer:

The density of water $1.00 g/ml=62.43 lbs/ft^3$ .

Explanation:

Density of water = 1.00 g/mL

1 lb = 453.592 g

$1 g=\frac{1}{453.592} lbs$

$ft^3=28316.8 mL$

$1 mL=\frac{1}{28316.8} ft^3$

Density of the water in $lb/ft^3$ :

$\frac{1.00 g}{1 mL}=\frac{1.00}{453.592 } lb\times \frac{28316.8 }{1 ft^3}$

$=62.43 lbs/ft^3$

The density of water $1.00 g/ml=62.43 lbs/ft^3$ .

3 0

4 years ago

Magbigay ng dahilan sa polusyon

Illusion [34]

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3 0

3 years ago

Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole

trapecia [35]

Answer : The enthalpy of combustion per mole of $C_6H_{12}O_6$ is -2815.8 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol$

Therefore, the enthalpy of combustion per mole of $C_6H_{12}O_6$ is -2815.8 kJ/mol

6 0

4 years ago

Read 2 more answers

What are two different types of electron microscopes and how might their ability to observe a specimen he different?

Sunny_sXe [5.5K]

Answer:

transmission EM (TEM) and the scanning EM (SEM)

Explanation:

Electron microscopy (EM) is a technique for obtaining high resolution images of biological and non-biological samples. It is used in biomedical research to examine the detailed structure of tissues, cells, organs and gross cell bodies.

There are two main type of electron microscope

the transmission EM (TEM)
the scanning EM (SEM)

Transmission electron microscopes are used to see thin specimens so that electrons can produce a projection image. TEM can be adapted in many ways to conventional (compound) light microscopes.
TEM is used for the internalization of cells , the formation of protein molecules, viruses, and organization of molecules in the cytoskeletal filaments. And arrangement of protein molecules in cell membranes.

4 0

3 years ago

Find the "natural" equilibrium concentration of NO in air in units of molecules/cm3 .

Vedmedyk [2.9K]

Answer:

6400 molecules / cm^3

Explanation:

10.6*10^-16 mol/L * 6.022*10^23 molecules/mol * 1 L / 1000 cm^3 * 1 / 100 L = 6400 molecules / cm^3

6 0

3 years ago