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2 years ago

Which family on the periodic table is least likely to enter into chemical reactions? 1 2 17 18

2 answers:

6 0

Elements present in group 18 are known as noble gases. The outermost shell of these elements are completely filled.

18 is the answer

finlep [7]2 years ago

5 0

Answer: Option (d) is the correct answer.

Explanation:

When an element has incompletely filled valence or outermost shell then that atom is most likely to undergo a chemical reaction in order to attain stability.

For example, sodium has atomic number 11 and its electronic configuration is 2, 8, 1. So, to attain stability sodium will readily loose 1 electron. Hence, it will become reactive in nature to attain stability.

Whereas argon has atomic number 18 and its electronic configuration is 2, 8, 8. Therefore, its valence shell is completely filled so, it will not react easily.

Hence, we can conclude that family of group 18 on the periodic table is least likely to enter into chemical reactions.

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You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain

IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of $O_2$ :-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{49.8\ g}{32\ g/mol}$

$Moles_{O_2}= 1.55625\ mol$

Since pressure and volume are constant, we can use the Avogadro's law as:-

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

Given ,

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

$\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}$

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of $CO_2$ = 44.0 g/mol

Mass of $CO_2$ = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

34.2 g is the mass of carbon dioxide gas one have in the container.

5 0

1 year ago

Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol

prohojiy [21]

Hey there!:

From the given data ;

Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar = g/L * mol/g

[E]t = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹

Hope that helps!

4 0

1 year ago

Missy Mae is given an unknown white ionic solid which contains one of the cations Na+, K+, Ca2+ and one of the anions Cl-, NO3-

topjm [15]

Answer:

NaNO₃

KNO₃

Ca(NO₃)₂

Explanation:

The possible cations for the salt. = Na⁺ , K⁺ , Ca²⁺

The anion for which the formula has to written = NO₃⁻

The salt is of white color.

The possible salts may be:

NaNO₃

KNO₃

Ca(NO₃)₂

Missy Mae can have one of these salts.

The color of NaNO₃ is from transparent to white.

The color of KNO₃ is white.

The color of NaNO₃ is from white to light grey.

7 0

2 years ago

his is the chemical formula for chromium(III) nitrate: . Calculate the mass percent of oxygen in chromium(III) nitrate. Round yo

Alinara [238K]

Answer:

$\%\ Composition\ of\ iron=69.92\ \%$

Explanation:

Percent composition is percentage by the mass of element present in the compound.

The formula for chromium(III) nitrate is $Cr(NO_3)_3$

Molar mass of chromium(III) nitrate = 238.011 g/mol

1 mole of chromium(III) nitrate contains 9 moles of oxygen

Molar mass of oxygen = 16 g/mol

So, Mass= Molar mass*Moles = 16*9 g = 144 g

$\%\ Composition\ of\ iron=\frac{Mass_{iron}}{Total\ mass}\times 100$

$\%\ Composition\ of\ iron=\frac{144}{238.011}\times 100$

$\%\ Composition\ of\ iron=69.92\ \%$

5 0

2 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

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

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

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

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

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

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

1 year ago