Green light has a frequency of 6.01*10^14 Hz. What is the wavelength?

Select all of the correct statements pertaining to the first and second ionization energies of different elements. check all tha

sertanlavr [38]

The First Ionization energy of Nitrogen is greater (Not smaller)than that of Phosphorous. This is because going down the group (N and P are in same group) the number of shells increases, the distance of valence electrons from Nucleus increases and hence due to less interaction between nucleus and valence electrons it becomes easy to knock out the electron.

The second ionization energy of Na is larger than that of Mg because after first loss of electron Na has gained Noble Gas Configuration (Stable Configuration) and now requires greater energy to loose both second electron and Noble Gas Configuration. While Mg after second ionization attains Noble Gas Configuration hence it prices less energy.

8 0

2 years ago

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Complete combustion of a compound containing hydrogen and carbon produced 2.641 g of carbon dioxide and 1.442 grams of water as

klio [65]

The empirical formula of hydrocarbon is C₃H₈

The molecular formula of hydrocarbon is C₆H₁₆

Empirical formula calculation

Hydrocarbon contain carbon and hydrogen

Step 1: find the mass carbon (C) in carbon dioxide (CO₂) and hydrogen (H ) in water

mass of of element = molar mass of element/ molar mass molecule x total mass of molecule

From periodic table the molar mass of C =12, for CO₂ = 12+( 16 x2) =44 g/mol, for H = 1.00 g/mol, for H₂O = (2 x1)+16 = 18 g/mol

mass of C = 12/44 x 2.641 =0.7203 g

since there are 2 atom of H in H₂O the molar mass of H = 1 x2 = 2 g/mol

mass of H is therefore = 2/18 x 1.442 =0.1602 g

Step 2: find the moles of C and H

moles = mass÷ molar mass

moles of C = 0.7203 g÷ 12 g/mol = 0.060 moles

moles of H = 0.1602÷ 1 g/mol = 0.1602 moles

Step 3: find the mole ratio of C and H by dividing each mole by smallest mole ( 0.06)

for C = 0.06/0.06 =1

For H = 0.1602/0.06 =2.67

multiply by 3 to remove the decimal

For C = 1 x3 =3

For H = 2.67 x3 =8

therefore the empirical formula = C₃H₈

The molecular formula calculation

[C₃H₈]n = 88.1 g/mol

[12 x 3)+( 1 x8)]n =88.1 g/mol

44 n = 88.1

divide both side by 44

n=2

therefore [C₃H₈]₂ = C₆H₁₆

7 0

1 year ago

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What is the molarity of 4.35 moles kmno4 dissolved in 750 ml of solution?

Andreyy89

M= #moles / L

4.35/.75 = 5.6

7 0

1 year ago

The mass unit associated with density is usually grams. if the volume (in ml or cm3) is multiplied by the density (g/ml or g/cm3

jok3333 [9.3K]

The weight in grams = 7.93 g

Given volume = 2.00 $in^{3}$

Given density = 0.242 g/ $cm^{3}$

We need to find the Mass(weight) in grams.

To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.

The unit of given volume is $in^{3}$ and unit of volume in density is $cm^{3}$ , so first we need to change the unit of volume from $in^{3}$ to $cm^{3}$ so that the volume units will cancel out, leaving only the mass units.

1 $in^{3}$ = 16.39 $cm^{3}$ (given conversion)

$2 in^{3}\times \frac{(16.39 cm^{3})}{(1 in^{3})}$

$in^{3}$ units get cancel out leaving the $cm^{3}$ unit.

$2 in^{3} = 32.78 cm^{3}$

Mass = Density X Volume.

Density = 0.242 g/ $cm^{3}$ and Volume = 32.78 $cm^{3}$

$Mass =0.242\frac{g}{cm^{3}}\times 32.78 cm^{3}$

Mass = 7.93 grams (g)

3 0

1 year ago

What is the percent composition by mass of nitrogen in the compound N2H4 (gram-formula mass = 32 g/mol)?

Alex

Answer:

$\large \boxed{93\, \% }$

Explanation:

1. Calculate the molar mass of N₂H₄

2N = 2 × 14 = 28

4H = 2 × 1 = 4

Tot. = 32

2. Calculate the mass percent of N

$\text{\% of element} = \dfrac{\text{mass of element}}{\text{mass of compound}} \times 100 \, \% = \dfrac{\text{28}}{\text{32}} \times 100 \, \% =\mathbf{88 \, \%}\\\\\text{The percentage of N in N$_{2}$H$_{4}$ is $\large \boxed{\mathbf{88\, \% }}$}}$

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2 years ago

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