6.579 rounded to nearest hundreth

Answers

Answer 1

Answer:

Answer:

6.58

Explanation:

Related Questions

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A cylinder is filled with 10.0L of gas and a piston is put into it. The initial pressure of the gas is measured to be 96.0kPa. The piston is now pulled up, expanding the gas, until the gas has a final volume of 45.0L. Calculate the final pressure of the gas. Be sure your answer has the correct number of significant digits.
A piston confines 0.200 mol Ne(g) in 1.20 at 25 degree C. Two experiments are performed. (a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm. (b) The gas is allowed to expand reversibly and isothermally to the same final volume. Please calculate the work done by the gas system in these two processes, respectively. Which process does more work? (revised from 6/e exercise 8.11) Please show calculation details.
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1. The common name for the compound CH3-CH2-O-CH3 is

James adds some magnetic marbles to a glass jar full of ordinary marbles, and then shakes up the jar.

Answers

Answer: Magnetic marbles will tend to attract each other

The heat capacity of chloroform (trichloromethane,CHCl3)in the range 240K to 330K is given
byCpm/(JK-1mol-1) = 91.47
+7.5x10-2(T/K). In a particular experiment,
1.0molCHCl3 is heated from 273K to 300K. Calculate the
changein molar entropy of the sample.

Answers

Answer : The change in molar entropy of the sample is 10.651 J/K.mol

Explanation :

To calculate the change in molar entropy we use the formula:

$\Delta S=n\int\limits^(T_f)_(T_i){(C_(p,m)dT)/(T)$

where,

$\Delta S$ = change in molar entropy

n = number of moles = 1.0 mol

$T_f$ = final temperature = 300 K

$T_i$ = initial temperature = 273 K

$C_(p,m)$ = heat capacity of chloroform = $91.47+7.5* 10^(-2)(T/K)$

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

$\Delta S=1.0\int\limits^(300)_(273){((91.47+7.5* 10^(-2)(T/K))dT)/(T)$

$\Delta S=1.0* [91.47\ln T+7.5* 10^(-2)T]^(300)_(273)$

$\Delta S=1.0* 91.47\ln ((T_f)/(T_i))+7.5* 10^(-2)(T_f-T_i)$

$\Delta S=1.0* 91.47\ln ((300)/(273))+7.5* 10^(-2)(300-273)$

$\Delta S=8.626+2.025$

$\Delta S=10.651J/K.mol$

Therefore, the change in molar entropy of the sample is 10.651 J/K.mol

Please help!!
which oneeeee

Answers

b is the correct answer

Determine the number of moles of oxygen atoms in each of the following.1) 4.93 mol H2O2
2) 2.01 mol N2O

Answers

Answer :

Part 1: 4.93 moles of $H_2O_2$ contains 9.86 moles of oxygen atoms.

Part 2: 2.01 moles of $N_2O$ contains 2.01 moles of oxygen atoms.

Explanation :

Part 1: 4.93 mol $H_2O_2$

In 1 mole of $H_2O_2$ , there are 2 atoms of hydrogen and 2 atoms of oxygen.

As, 1 mole of $H_2O_2$ contains 2 moles of oxygen atoms.

So, 4.93 moles of $H_2O_2$ contains $4.93* 2=9.86$ moles of oxygen atoms.

Thus, 4.93 moles of $H_2O_2$ contains 9.86 moles of oxygen atoms.

Part 2: 2.01 mol $N_2O$

In 1 mole of $N_2O$ , there are 2 atoms of nitrogen and 1 atom of oxygen.

As, 1 mole of $N_2O$ contains 1 mole of oxygen atoms.

So, 2.01 moles of $N_2O$ contains $2.01* 1=2.01$ moles of oxygen atoms.

Thus, 2.01 moles of $N_2O$ contains 2.01 moles of oxygen atoms.

Calcium and bromine have formed a bond. Leading up to this, calcium gave up electrons. It was a(n)

Answers

Leading up to this, calcium gave up 2 valence electrons and thus was denoted as a cation. These 2 electrons were transferred to bromine, which received an overall negative charge because of the addition of 2 valence electrons in its valence shell, and thus formed a negatively charged ion, an anion.

Both formed an ionic bond, due to the electrostatic charge of attraction between the 2 oppositely charged ions. If many ions of Ca and Br are present and numerous ionic bonds have formed it will undergo an arrangement which is that of an ionic lattice, type of structure.

Which of the following correctly identifies which has the higher first ionization energy, CI or Ar, and supplies the best justification? a. Cl, because of its higher electronegativity b. Cl, because of its higher electron affinity c. Ar, because of its completely filled valence shell d. Ar, because of its higher effective nuclear charge Given the information in a table of bond dissociation energies, calculate the change in enthalpy, delta H, in units of kJ, for the following gas-phase reaction: H_2C = CH_2 + H-Br rightarrow CH_3CH_2Br a. +148 b. -148 c. +200 d. -200

Answers

Answer:

d. Ar, because of its higher effective nuclear charge

For the secon part see explanation below.

Explanation:

The first ionization energy is the energy required to remove an electron from the atom from its outermost shell. It depends on the nuclear charge, distance from the nucleus and the screening of other electrons in the inner shells of the atom.

Comparing Cl and Ar we see that being both elements of the third period, the Ar atom has one more proton than Cl and therefore the electron feels more nuclear charge making the first ionization of Ar greater than Cl.

a) False, electronegativity relates to attraction for an electron and not to the first ionization.

b) False, again electron affinity is not first ionization, it is defined as the energy released when the atom captures an added electron.

c) False,athough it is true that Ar has a complete octet, the higher first ionization is affected by nuclear charge. The screening of electrons in the n= 1 and 2 shells is almost the same so what is important is that the electrons in the n= 3 shell feel more nuclear charge.

d) True for all the reasons given previously : the higher effective nuclear charge in Ar.

For the second part, we have to make an inventory of the bonds being broken and formed:

ΔHºrxn = H broken - H formed, where H is the bond energy

H2 C = CH_2 + H-Br ⇒ CH_3CH_2Br

ΔHºrxn = ( 1 C=C + 4 C-H + 1 H-Br) - ( 1 C-C + 5 C-H + 1 C-Br)

ΔHºrxn (kJ) = (614 + 4(413) + 363) - ( 347 + 5 (413) + 276)

ΔHºrxn (kJ) = 2629 - 2688 = -59 kJ

This value is not in the choices due to mistaken bond energy values from the tables.

Answer:

1. Ar, because of its higher effective nuclear charge.

2. ∆Hrxn = -200 KJ/mol

Explanation:

The size of the atoms of chemical elements can be measured from their atomic radius which is also affected by the effective nuclear charge.

Recall that elements in a particular period have the same number of electron shells. Also, along a given period, atomic radius decreases due to an increase in the effective (positive) nuclear charge. This is because as the atomic (proton) number increases along that period, the charge on the nucleus also increases. With more protons in the nucleus the overall attraction between the positively charged nucleus and the negatively charged surrounding electrons increases, so the electrons are pulled closer to the nucleus thereby leading to a decrease in the atomic size.

So, along a given period atomic size decreases due to an increase in the effective nuclear charge.

The first ionization energy is the minimum energy (in kilojoules) needed to strip one mole of electrons from one mole of a gaseous atom of an element to form one mole of a gaseous unipositive ion.

Along a particular period, ionization energy increases due to an increase in the effective nuclear charge and a decrease in atomic radius. This is because, the smaller the atom the more stable it is and the more difficult it will be to remove an electron.

For the second question,

The enthalpy change of a reaction is the difference in the bond dissociation energies of the reactants and products. Bonds are broken in reactant molecules and formed in product molecules. Bond breaking energies are usually intrinsic ( endothermic, +be ∆H ) while bond forming energies are usually extrinsic ( exothermic, -ve ∆H ).

So,

∆Hrxn = n∆H(reactants/bonds broken) - m∆H(products/bonds formed)

Where n and m = stoichiometric coefficients of the products and reactants respectively from the balanced chemical equation.

First, draw the correct Lewis structures of the compounds.

Next, identify all the bonds broken and formed.

Then, from the bond dissociation energies ( usually given or looked up in texts ), sum up the bond breaking energies and the bond forming energies and subtract the bond forming energies from the bond breaking energies.

Considering this equation:

H_2C = CH_2 + H-Br rightarrow CH_3CH_2Br

The equation is balanced.

Bonds broken (number of bonds ):

I. C=C (1)

II. H-Br (1)

III. C-H (4)

Bonds formed:

I. C-C (1)

II. C-H (5)

III. C-Br (1)

∆Hrxn = [ ( 1 x C=C ) + ( 4 x C-H ) + ( 1 x H-Br ) ] – [ ( 1 x C-C ) + ( 5 x C-H ) + ( 1 x C-Br ) ]