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3 years ago

What is an alkali metal with fewer than 10 protons in its nucleus?

Chemistry

1 answer:

Trava [24]3 years ago

6 0

Answer:

Lithium

Explanation:

Alkali metals are group of metals which are present in first group of periodic table. As we know atomic number is equal to number of protons contained by a particular element. Therefore, the alkali metals along with there number of protons are listed below;

Alkali Metal Number of Protons

Lithium 3

Sodium 11

Potassium 19

Rubidium 37

Cesium 55

Francium 87

Hence, it is cleared from above table that Lithium is having fewer protons than 10.

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Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota

Ghella [55]

Answer:

(a) Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = -2.881 J/K; total change in entropy = 0

(b)Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = 0 ; total change in entropy = 2.881 J/K

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = $V_{1}$

Final volume = $V_{2}$ = $2V_{1}$

(a) Change in entropy of the system Δ $S_{sys}$ = $nRIn\frac{V_{2} }{V_{1} }$

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

Δ $S_{sys}$ = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding Δ $S_{sur}$ = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ $S_{sys}$ = 2.881 J/K

Since surrounding does not change in this process Δ $S_{sur}$ = 0.

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 0 = 2.88 J/K

Δ $S_{sys}$ = 0

Since heat energy is not transferred from the system to the surrounding

Δ $S_{sur}$ = 0

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 0

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3 years ago

The CNO cycle produces energy in which of the following?

d1i1m1o1n [39]

Stars on the main sequence fuse hydrogen into helium via a six-stage sequence of reactions

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3 years ago

Which portion of a molecule of F2O has partial positive charge?

german

Answer:

Explanation:

I hope this may help you guys

6 0

2 years ago

Use your understanding of the ideal gas law to

PolarNik [594]

Answer:The ideal gas law is represented mathematically as: PV=nRT. P- pressure, V- volume, n-number of moles of gas, R- ideal gas constant, T- temperature.

Explanation:The ideal gas law is used as a prediction of the behavior of many gases, when subjected to different conditions.

he ideal gas law has so many limitations.

An increase in the pressure or volume, decreases the number of moles and temperature of the gas.

Empirical laws that led to generation of the ideal gas laws, considered two variables and keeping the others constant. This empirical laws include, Boyle's law, Charles's law, Gay Lusaac's law and Avogadro's law.

4 0

3 years ago

Read 2 more answers

If the two atoms participating in a bond have an electron negativity difference of 1.9 , what type of of bond would they be part

brilliants [131]

Answer:

Polar covalent bond.

Explanation:

When the bond is formed between the atoms by sharing the electrons the bond thus have covalent character. The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive. When the electronegativity difference is less than 0.4 the bond is non polar covalent.

When bonded atoms have greater electronegativity difference i.e 2 or greater than two the bond is ionic because electron is transfer from low electronegative atom to highest electronegative atom.

For example:

In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.

In case of H₂, Cl₂, Br₂ the bond has very high covalent character because of zero electronegativity difference.

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3 years ago