How many GRAMS are in 3.78 mol of NaOH? A. 150 g/mol B. 151.12 g/mol C. 151.1 g/mol

Is work done whenever you hold a heavy object for a long time?

AnnZ [28]

No, work is not done whenever you hold a heavy object for a long time

The result of a force's displacement and its component of force exerted by the object in the direction of displacement is what is known as the force's work. When we push a block with some force, the body moves quickly and work is completed.

No work, as that term is used here, is done until the object is moved in some way and a component of the force travels along the path that the object is moved. Because there is no displacement when holding a heavy object still, energy is not transferred to it.

Learn more about Work done here:

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7 0

1 year ago

Where are the noble gases on the periodic table and what characteristics do they have?

OlgaM077 [116]

the noble gases are those who occupy the eighth group of the periodic table and are so called because having the complete valence shell do not need to form bonds with other atoms and are thus in atomic form ...

The noble gases (also called rare gases) are of the inert gases that constitute the eighteenth [1] group of the periodic table of the elements, ie, the right-most column. They consist of atoms with electron shells full. It includes the following elements: helium neon argon krypton xenon radon Ununoctium Sometimes they (particularly helium) are located together with other gases (mostly nitrogen and methane) into endogenous sources; helium of endogenous origin comes from the decomposition of radioactive elements present in the subsurface that emit α particles (ie ions He2 +): These oxidized species present in the soil and become elio.I atoms of the noble gases are all monatomic gas, not easily liquefiable, present the atmosphere in different percentages; the most common is argon which is approximately the 0.932%.

4 0

3 years ago

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15.0 moles of gas are in a 8.00 L tank at 22.3 ∘C∘C . Calculate the difference in pressure between methane and an ideal gas unde

leva [86]

Answer:

$\Delta P=4.10atm$

Explanation:

Hello!

In this case, since the ideal gas equation is used under the assumption of no interaction between molecules and perfectly sphere-shaped molecules but the van der Waals equation actually includes those effects, we can compute each pressure as shown below, considering the temperature in kelvins (22.3+273.15=295.45K):

$P^{ideal}=\frac{nRT}{V}=\frac{15.0mol*0.08206\frac{atm*L}{mol*K}*295.45K}{8.00L}=45.5atm$

Next, since the VdW equation requires the molar volume, we proceed as shown below:

$v=\frac{8.00L}{15.0mol}=0.533\frac{L}{mol}$

Now, we use its definition:

$P^{VdW}=\frac{RT}{v-b} -\frac{a}{v^2}$

Thus, by plugging in we obtain:

$P^{VdW}=\frac{0.08206\frac{atm*L}{mol*K}*295.45K}{0.533mol/L-0.0430L/mol} -\frac{2.300L^2*atm/mol^2}{(0.533L/mol)^2}\\\\P^{VdW}=49.44atm-8.09atm\\\\P^{VdW}=41.4atm$

Thus, the pressure difference is:

$\Delta P=45.5atm-41.4atm\\\\\Delta P=4.10atm$

Best regards!

6 0

3 years ago

Glucose (C6H12O6) is a key nutrient for generating chemical potential energy in biological systems. We were provided 16.55 g of

harina [27]

Answer:

a) 40 %

b) $4.04~g~CO_2$

c) $5.53x10^2^3~molecules~of~O_2$

Explanation:

For a) we will have to calculate the molar mass of $C_6H_1_2O_6$ , so the first step is to find the atomic mass of each atom and multiply by the amount of atoms in the molecule.

C => 12*(6) = 72

H => 1*(12) = 12

O => 6*(16) = 96

Molar mass = 180 g/mol

Then we can calculate the percentage by mass:

$Percentage~=~\frac{72}{180}*100=40$

For b) we have to start with the reaction of glucose:

$C_6H_1_2O_6~+~6O_2~->~6CO_2~+~6H_2O$

Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the following conversion ratios:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol carbon dioxide

-) 1 mol carbon dioxide = 44 g carbon dioxide

$16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~CO_2}{1~mol~C_6H_1_2O_6}\frac{44~g~CO_2}{1~mol~CO_2}=4.04~g~CO_2$

For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the following conversion ratios:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol oxygen

-) 1 mol oxygen = 6.023x10^23 molecules of O2

$16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~O_2}{1~mol~C_6H_1_2O_6}\frac{6.023x10^2^3~molecules~O_2}{1~mol~O_2}=~5.53x10^2^3~molecules~of~O_2$

4 0

2 years ago

Which pair shares the same empirical formula?

MissTica

Answer:- 3. $CH_3$ and $C_2H_6$

Explanations:- An empirical formula is the simplest whole number ratio of the moles of atoms present in the compound.

If we simplify the first par formula then they becomes $CH_2$ and CH. This pair does not share the same empirical formula.

Second pair formulas are already simple and could not be further simplify. Since they are different, it is also not the right option.

Fourth pair has CH and $C_2H_4$ . $C_2H_4$ could be simplify to $CH_2$ . The two formulas are still different and so it is not the right choice.

The only correct one is third pair as one of them is $CH_3$ and other one is $C_2H_6[/tex . [tex]C_2H_6$ could be simplify to $CH_3$ . This pair share the same empirical formula, $CH_3$ .

6 0

3 years ago