9

Which of the following pieces of information about the element lithium describes the atomic number only

vagabundo [1.1K]

The top left corner shows the atomic number which is 3.

6 0

3 years ago

A 100-watt light bulb radiates energy at a rate of 100 J/s. (The watt, a unit of power or energy over time, is defined as 1 J/s.

Semmy [17]

Answer

2.7956 * 10^19 photons

Givens

Wavelength = λ = 525 * 10^-9 meters [1 nmeter = 1*10^-9 meters]
c = 3 * 10^8 meters
E = ???
W = 100 watts
t = 1 second
h= plank's Constant = 6.26 * 10^-34 J*s

Formula

E = h * c / λ

W = E / t

Solution

E = 6.26 * 10^-34 j*s * 3 * 10^8 m/s /525 * 10^-9 (m)

The meters cancel out. So do the seconds. You are left with Joules as you should be.

E = 3.577 * 10^-18 Joules

What you have found is the energy of 1 photon.

Now you have to find the Joules from the watts.

W = E/t

100 * 1 second = 100 joules

1 photon contains 3.577 * 10 ^ - 18 Joules

x photon = 100 joules

1/x = 3.577 * 10^-18 / 100 Cross multiply

100 = 3.577 * 10 ^ - 18 * x Divide both sides by 3.577 * 10 ^ - 18

100/3.577 * 10 ^ - 18 = 3.577 * 10 ^ - 18x / 3.577 * 10 ^ - 18

2.7956 * 10^19 photons = x

7 0

3 years ago

Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

$\Delta H = \Delta H_{neut} - \Delta H_{soln}$

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

brainly.com/question/2082986?referrer=searchResults

brainly.com/question/1657608?referrer=searchResults

I hope it helps you!

6 0

2 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

3 years ago

Calculate the mass of ethyl alcohol required to prepare 540 grams of C4H6, if the reaction follows the scheme:

kkurt [141]

Answer:

Explanation:

2C₂H₅OH = C₄H₆ + 2H₂O + H₂

2 mole 1 mole

molecular weight of ethyl alcohol

mol weight of C₂H₅OH = 46 gm

mol weight of C₄H₆ 54 gm

540 gm of C₄H₆ = 10 mole

10 mole of C₄H₆ will require 20 mol of ethyl alcohol .

20 mole of ethyl alcohol = 20 x 46

= 920 gm

ethyl alcohol required = 920 gm .

3 0

3 years ago