The equation for determining the energy of a photon of electromagnetic radiation is #E=hnu# , where E is energy in Joules, h is Planck's constant, #6.626 xx 10^(-34)J*s# , and #nu# (pronounced "nu") is the frequency

hope this helps

8 0

3 years ago

What intermolecular forces are present in each of the substances?

Nesterboy [21]

What intermolecular forces are present in each of the substances? CH4,C3H8,CH3F,HF, C6H5OH (dispersion forces, dipole-dipole forces, or hydrogen bonding);A sample of ideal gas at room temperature occupies a volume of 34.0L at a pressure of 782torr .

4 0

3 years ago

Read 2 more answers

What volume of H2O(g) is produced when 8.00 mol of C2H4(g) reacts at STP?

sweet-ann [11.9K]

* Mole ratio:

C2H4 + 3 O2= 2 CO2 + 2 H2O

1 mole C2H4 --------------- 2 moles H2O
8.00 moles C2H4 ---------- ?

8.00 x 2 / 1 => 16 moles of H2O

Therefore:

1 mole --------- 22.4 L at ( STP)
16 moles ------- ?

16 x 22.4 / 1 => 358.4 L

hope this helps!

4 0

3 years ago

When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

3 0

4 years ago