Someone please help me

Which of the following is a correct description of a double bond between two carbon atoms

kicyunya [14]

A double bond between carbon atoms is longer than a triple bond between carbon atoms.

5 0

3 years ago

Choose the products that complete the reaction . The chemical equation may not be balanced . Al+H 2 SO 4 ? Al 2 (SO 4 ) 3 +H 2 O

alexandr402 [8]

Answer: Al2 (SO4)3 + H2

Explanation:

option C

5 0

3 years ago

Use the Nernst equation to calculate the concentration of the unknown solution. Base this on your experimental voltage of 1.074

Hoochie [10]

Answer:

0.3793 M

Explanation:

The unknown metal is zinc. So the equation of the reaction is;

Zn(s) + Cu^2+(aq) -------> Zn^2+(aq) + Cu(s)

From Nernst equation;

E = E° - 0.0592/n log Q

[Cu2+] = 0.050179 M

n = 2

[Zn^2+] = ?

E = 1.074 V

E° = 0.34 - (-0.76) = 1.1 V

Substituting values;

1.074 = 1.1 - 0.0592/2 log [Zn^2+]/0.050179

1.074 - 1.1 = - 0.0592/2 log [Zn^2+]/0.050179

-0.026 = -0.0296 log [Zn^2+]/0.050179

-0.026/-0.0296 = log [Zn^2+]/0.050179

0.8784 =log [Zn^2+]/0.050179

Antilog(0.8784) = [Zn^2+]/0.050179

7.558 = [Zn^2+]/0.050179

[Zn^2+] = 7.558 * 0.050179

[Zn^2+] = 0.3793 M

4 0

3 years ago

Sometimes in lab we collect the gas formed by a chemical reaction over water (see sketch at right). This makes it easy to isolat

zepelin [54]

Answer:

The correct answer is 0.12 grams.

Explanation:

The mass of carbon monoxide or CO collected in the tube can be determined by using the ideal gas equation, that is, PV = nRT.

Based on the given question, P or the pressure of the gas is given as 1 atm, volume of the gas collected in the tube is 117 ml or 0.117 L.

The number of moles or n can be determined by using the equation, mass/molar mass.

R is the universal gas constant, whose value is 0.0821 L atmK^-1mol^-1, and temperature is 55 degree C or 328 K (55+273).

On putting the values we get:

n = PV/RT

= (1 atm*0.117 L) / (0.0821 L atmK^-1mol^-1 * 328 K)

= 0.0043447 mol

Therefore, mass of CO will be moles * molar mass of CO

= 0.0043447 mol * 28 g/mol

= 0.12 g

3 0

3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

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

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

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

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago