Chem help 36 and 37 the word cut off on 37 is larger

Katarina [22]

Answer:

single bond= 1 sigma bond

so basically 1

Explanation:

Usually, all bonds between atoms in most organic compounds contain one sigma bond each. If it is a single bond, it contains only sigma bond. Multiple bonds (double and triple), however, contains sigma and pi bonds. Double bonds have one each, and triple bonds have one sigma bond and two pi bonds

8 0

3 years ago

Read 2 more answers

Approximately how many ice cubes must melt to cool 650 milliliters of water from 29°C to 0°C? Assume that each ice cube contains

qwelly [4]

Answer : The number of ice cubes melt must be, 13

Explanation :

First we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}$

Density of water = 1.00 g/mL

Volume of water = 650 mL

$\text{Mass of water}=1.00g/mL\times 650mL=650g$

Now we have to calculate the heat released on cooling.

Heat released on cooling = $m\times c\times (T_2-T_1)$

where,

m = mass of water = 650 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_2$ = final temperature = $29^oC$

$T_2$ = initial temperature = $0^oC$

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

Heat released on cooling = $650g\times 4.18J/g^oC\times (29-0)^oC$

Heat released on cooling = 78793 J = 78.793 kJ (1 J = 0.001 kJ)

As, 1 ice cube contains 1 mole of water.

The heat required for 1 ice cube to melt = 6.02 kJ

Now we have to calculate the number of ice cubes melted.

Number of ice cubes melted = $\frac{\text{Total heat}}{\text{Heat for 1 ice cube}}$

Number of ice cubes melted = $\frac{78.793kJ}{6.02kJ}$

Number of ice cubes melted = 13.1 ≈ 13

Therefore, the number of ice cubes melt must be, 13

3 0

3 years ago

Calcule la normalidad de una solución

tensa zangetsu [6.8K]

Answer:

si por favor

Explanation:

5 0

3 years ago

In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t

solmaris [256]

Answer:

4.81 moles

Explanation:

The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.

Pressure at which gauge reads zero = 14.7 psi

Pressure read by the gauge = 988 psi

Total pressure = 14.7 + 988 psi = 1002.7 psi

Also, P (psi) = P (atm) / 14.696

Pressure = 1002.7 / 14.696 = 68.2297 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 4.81 moles

4 0

3 years ago

Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f

Zanzabum

K = 5.0 × 10^25

<h2>Part (a). Calculate E° for the reaction </h2>

Step 1. Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

Step 2. Identify the cathode and the anode

The half-cell with the more negative E° (Zn) is the anode.

Step 3. Calculate E°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

E° = +0.76 V

<h2>Part (b). Calculate K for the reaction </h2>

The relation between E° and K is

E° = (RT)/(nF)lnK

where

R = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

T = the Kelvin temperature

n = the moles of electrons transferred

F = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]lnK

0.76 = 0.012 85 lnK

lnK = 0.76/0.012 85 = 59.16

K =e^59.16 = 5.0 × 10^25

4 0

3 years ago

Chem help 36 and 37 the word cut off on 37 is larger​

Chem help 36 and 37 the word cut off on 37 is larger