G.com what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume

Explanation:

According to the mole concept, there are [tex]6.022 \times 10^{22}[/tex] atoms present.

It is given that mass is 0.250 g. And, number of moles are equal to mass divided by molar mass.

Mathematically,       No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

As molar mass of [tex]P_{2}O_{5}[/tex] is 283.88 g/mol. Therefore, putting given values into the above formula as follows.

                 No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                                       = [tex]\frac{0.250 g}{283.88 g/mol}[/tex]

                                       = 0.0008 mol

Hence, number of atoms present in 0.0008 mol are as follows.

                      [tex]0.0008 mol \times 6.022 \times 10^{22} atoms/mol[/tex]

                     = 0.0048 atoms

Thus, we can conclude that there are 0.0048 atoms in 0.250 g of [tex]P_{2}O_{5}[/tex].

Answer:

If we have 29 quibs of skvarnick they will be equal to 11.6 sleps.

Explanation:

Mass of a substance called skvarnick = 29 quibs

10 quibs is equal to 4 sleps. This means that 1 quibs is equal to 0.4 sleps.

10 quibs = 4 sleps

1 quibs = [tex]\frac{4}{10} sleps = 0.4 sleps[/tex]

Then 29 quibs will be:

[tex]29 quibs=29\times 0.4 sleps =11.6 sleps[/tex]

If we have 29 quibs of skvarnick they will be equal to 11.6 sleps.

Answer : The correct option is, (d) He and Be

Explanation :

First we have to determine the electronic configuration of the following elements.

The electronic configuration of hydrogen (H) is:

[tex]1s^1[/tex]

The electronic configuration of carbon (C) is:

[tex]1s^22s^22p^2[/tex]

The electronic configuration of lithium (Li) is:

[tex]1s^22s^1[/tex]

The electronic configuration of helium (He) is:

[tex]1s^2[/tex]

The electronic configuration of beryllium (Be) is:

[tex]1s^22s^2[/tex]

From the electronic configuration of the elements we conclude that, the hydrogen element has '1' valence electron, carbon element has '4' valence electrons, lithium element has '1' valence electron, helium element has '2' valence electrons and beryllium element has '2' valence electrons.

Thus, the helium and beryllium are the elements that holds up to two electrons in their outermost shell.

Hence, the correct option is, (d) He and Be

Answer: The balanced chemical equation is written below.

Explanation:

Balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side. These equations follow law of conservation of mass.

The chemical equation for the reaction of lead (II) nitrate and potassium phosphate follows:

[tex]3Pb(NO_3)_2(aq.)+2K_3PO_4(aq.)\rightarrow Pb_3(PO_4)_2(s)+6KNO_3(aq.)[/tex]

By Stoichiometry of the reaction:

3 moles of aqueous lead (II) nitrate reacts with 2 moles of aqueous solution of potassium phosphate to produce 1 mole of solid lead (II) phosphate and 6 moles of aqueous solution of potassium nitrate.

Hence, the balanced chemical equation is written above.

Two constitutional isomers exist for C2H5Cl: 1-chloroethane (ethyl chloride) and 2-chloroethane (chloroethane), with chlorine bonded to different carbon atoms in each isomer.

The student has asked to draw structures for all constitutional isomers with the molecular formula C2H5Cl, which is a chemical exercise focusing on understanding the different ways in which atoms can be rearranged in space to create molecules with the same molecular formula but different structures.

There are actually only two constitutional isomers with the formula C2H5Cl. These isomers are:

In these structures, the chlorine atom is bonded to different carbon atoms, resulting in molecules that have different physical and chemical properties.

Final answer:

To find out how many 0.05 mL drops of blood are in a 2 mL blood collection tube, divide the total volume by the volume of one drop. The calculation shows that there are 40 drops in a 2 mL tube.

Explanation:

To determine how many drops of blood are in a blood collection tube that holds 2 mL, we need to understand the relationship between the volume of the drops and the total volume that the tube can hold. Given that each drop of blood is 0.05 mL, we can calculate the number of drops in 2 mL by dividing the total volume by the volume of a single drop.

Here is the step-by-step calculation:

Determine the volume of one drop: 0.05 mL.

Determine the total volume of the collection tube: 2 mL.

Divide the total volume by the volume of one drop: 2 mL / 0.05 mL per drop.

Calculate the number of drops: 40 drops.

The two Lewis structure for a compound with the formula [tex]\rm C_4H_{10}[/tex] are shown below in the attached image.

In a Lewis structure, an element's chemical symbol shows its nucleus and inner electrons, while dots or lines represent valence electrons. Valence electrons are the atom's outermost electrons and are important in chemical bonding.

Therefore, the two structure of [tex]\rm C_4H_{10}[/tex] are isomers of each other (refer to the attached image).

Learn more about Lewis structure here:

https://brainly.com/question/9908915

#SPJ6

Answer:

6 electrons are in the third level if the atom.

Answer:

The degree of rotation of racemic mixture will equal 0° 

Explanation:

Hello,

(-) isomer is in other words the Enantiomer of (+) isomer and as long as the statement indicates that (+)-isomer optical rotation is + 13.5°, the optical rotation of (-)-isomer will be the same value and degree but with the inverse sign which is -13.5°.

In such a way, the degree of rotation of the taken into account racemic mixture will equal 0° 

This is in this way due to the fact that the racemic mixture has equal amount of both enantiomers.

Best regards.

Final answer:

The atomic mass closest to carbon without calculations is 12.00 amu since carbon's average atomic mass is listed as 12.011 amu on the periodic table, and it mostly consists of carbon-12.

Explanation:

The atomic mass of an element like carbon is the weighted average of the masses of its naturally occurring isotopes. Carbon is predominantly composed of carbon-12, which has an atomic mass of exactly 12 amu (atomic mass units), as this isotope makes up about 98.90% of naturally occurring carbon. Because of this, the average atomic mass of carbon is close to 12 amu, but slightly higher due to the presence of the carbon-13 isotope. The periodic table lists carbon's atomic mass as 12.011 amu, making 12.00 amu the mass closest to carbon's atomic mass without doing any calculations.

Final answer:

The substance with a molar mass of 44.01 g/mol is carbon dioxide (CO2), which can be deduced by calculating the combined molar masses of one carbon atom and two oxygen atoms.

Explanation:

The question concerns the identification of a compound with a molar mass of 44.01 g/mol. One well-known substance with this molar mass is carbon dioxide (CO2). To arrive at this conclusion, one can examine the atomic masses of carbon and oxygen. Carbon has a molar mass of about 12.01 g/mol, while oxygen has a molar mass of about 16.00 g/mol. Given that carbon dioxide is composed of one carbon atom and two oxygen atoms, the total molar mass can be calculated as follows: 12.01 amu (for carbon) + 2 × 16.00 amu (for each oxygen) = 44.01 amu, which corresponds to g/mol when talking about molar mass. Hence, the molecular mass of carbon dioxide is 44.01 g/mol.

n = 6 to n = 2

From several sources, we have prepared the following answer choices:

A. n = 2 to n = 5  

B. n = 6 to n = 4  

C. n = 3 to n = 2  

D. n = 6 to n = 2  

We will determine which electron transition would correspond to the shortest wavelength line in the visible emission spectra for hydrogen.  

The amount of energy released or absorbed by electrons when moving from n₁ level to n₂ level is equal to  

[tex]\boxed{ \ \Delta E = -13.6 \Big( \frac{1}{n_2^2} - \frac{1}{n_1^2} \Big) \ }[/tex]  in eV.

This energy difference is equal to [tex]\boxed{ \ hf = \frac{hc}{\lambda} \ }[/tex], where f and λ are the frequency and wavelength of the radiation emitted or absorbed.

Thus, the wavelength is inversely proportional to the energy difference from the electron transition. To get the shortest wavelength, it is determined by the largest ΔE.  

From the formula above, we practically only need to calculate part [tex]\boxed{ \ \Big( \frac{1}{n_2^2} - \frac{1}{n_1^2} \Big) \ }[/tex] which is directly proportional to ΔE.  Then from the results of the calculation of this section, we will get the shortest wavelength from the largest result..

A. n₁ = 2 to n₂ = 5  

[tex]\boxed{ \ \Big( \frac{1}{5^2} - \frac{1}{2^2} \Big) \ }[/tex]  

[tex]\boxed{ \ -\frac{21}{100} \ }[/tex]

By taking the absolute value, we get [tex]\boxed{ \ 0.210 \ }[/tex]

B. n₁ = 6 to n₂ = 4  

[tex]\boxed{ \ \Big( \frac{1}{4^2} - \frac{1}{6^2} \Big) \ }[/tex]  

[tex]\boxed{ \ \frac{5}{144} \ }[/tex]

We get [tex]\boxed{ \ 0.0347 \ }[/tex]

C. n₁ = 3 to n₂ = 2  

[tex]\boxed{ \ \Big( \frac{1}{2^2} - \frac{1}{3^2} \Big) \ }[/tex]  

[tex]\boxed{ \ \frac{5}{36} \ }[/tex]

We get [tex]\boxed{ \ 0.1389 \ }[/tex]

D. n₁ = 6 to n₂ = 2  

[tex]\boxed{ \ \Big( \frac{1}{2^2} - \frac{1}{6^2} \Big) \ }[/tex]  

[tex]\boxed{ \ \frac{2}{9} \ }[/tex]

We get [tex]\boxed{ \ 0.222 \ }[/tex]

The last calculation above shows the greatest results so that the shortest wavelength is undoubtedly gained from the electron transition n = 6 to n = 2.

Keywords: according to the Bohr model of the atom, which electron transition would correspond, to the shortest wavelength, the visible emission spectra for hydrogen, energy, inversely proportional

The electronic transition from [tex]\boxed{{\text{D}}{\text{. n}} = {\text{6 to n}} = {\text{2}}}[/tex] corresponds to the shortest wavelength.

Further explanation:

Rydberg equation describes the relation of wavelength of spectral line with the transition values. The expression for Rydberg equation is as follows:

[tex]\dfrac{1}{\lambda } = \left( {{{\text{R}}_{\text{H}}}} \right)\left( {\dfrac{1}{{{{\left( {{{\text{n}}_{\text{1}}}} \right)}^2}}} - \dfrac{1}{{{{\left( {{{\text{n}}_{\text{2}}}} \right)}^2}}}} \right)[/tex]  …… (1)                                                  

Here,

[tex]\lambda[/tex] is the wavelength of spectral line

[tex]{{\text{R}}_{\text{H}}}[/tex] is Rydberg constant that has the value  

[tex]{{\text{n}}_{\text{1}}}[/tex] and [tex]{{\text{n}}_{\text{2}}}[/tex] are the two positive integers, where  .

Rearrange equation (1) to calculate  .

[tex]\lambda = \dfrac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left( {\dfrac{1}{{{{\left( {{{\text{n}}_1}} \right)}^2}}} - \dfrac{1}{{{{\left( {{{\text{n}}_{\text{2}}}} \right)}^2}}}} \right)}}[/tex]                                        …… (2)

A. n = 2 to n = 5

Substitute 2 for [tex]{{\text{n}}_{\text{1}}}[/tex] and 5 for [tex]{{\text{n}}_{\text{2}}}[/tex] in equation (2).

 [tex]\begin{aligned}\lambda&= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left( {\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 5 \right)}^2}}}} \right)}} \\&= 4.34 \times {10^{ - 7}}{\text{ m}}\\\end{aligned}[/tex]

B. n = 6 to n = 4

Substitute 4 for [tex]{{\text{n}}_{\text{1}}}[/tex] and 6 for [tex]{{\text{n}}_{\text{2}}}[/tex] in equation (2).

 [tex]\begin{aligned}\lambda&= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left( {\frac{1}{{{{\left( 4 \right)}^2}}} - \frac{1}{{{{\left( 6 \right)}^2}}}} \right)}}\\&= 2.63 \times {10^{ - 6}}{\text{ m}}\\\end{aligned}[/tex]

C. n = 3 to n = 2

Substitute 2 for [tex]{{\text{n}}_{\text{1}}}[/tex] and 3 for [tex]{{\text{n}}_{\text{2}}}[/tex] in equation (2).

 [tex]\begin{aligned}\lambda  &= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left( {\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 3 \right)}^2}}}} \right)}} \\ &= 6.56 \times {10^{ - 7}}{\text{ m}}\\\end{aligned}[/tex]

D. n = 6 to n = 2

Substitute 2 for [tex]{{\text{n}}_{\text{1}}}[/tex] and 6 for [tex]{{\text{n}}_{\text{2}}}[/tex] in equation (2).

 [tex]\begin{aligned}\lambda&= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left({\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 6 \right)}^2}}}}\right)}} \\&= 4.10 \times {10^{ - 7}}{\text{ m}}\\\end{aligned}[/tex]

The value of [tex]\lambda[/tex] for transition from n = 6 to n = 2 is the least and therefore this transition corresponds to the shortest wavelength.

Learn more:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: Rydberg constant, wavelength, n1, n2, positive integers, transition, 2, 6, 3, 5, transition, Rh, spectral line, shortest wavelength.

The atomic mass of rubidium can be determined using the mass spectrum of rubidium obtained from a mass spectrometer.

The atomic mass of rubidium can be determined using the mass spectrum of rubidium obtained from a mass spectrometer. In a mass spectrometer, a sample of rubidium is vaporized and exposed to high-energy electrons, causing the rubidium atoms to become charged ions. These ions are then accelerated into a magnetic field, and the extent to which they are deflected depends on their mass-to-charge ratios. By measuring the relative deflections of the ions and analyzing the mass spectrum, chemists can determine the mass of rubidium.

To take the percent by mass of this element, we use the formula:

% mass = (mass of element / mass of ore) * 100%

% mass = (47.5 g / (660 kg * 1000 g / kg)) * 100*

% mass = 7.20 x 10^-3 %

Answer: The mass percent of element in the ore is 0.0072 %

Explanation:

To calculate the mass percentage of element in ore, we use the equation:

[tex]\text{Mass percent of element}=\frac{\text{Mass of element}}{\text{Mass of ore}}\times 100[/tex]

Mass of element = 47.5 g

Mass of ore = 660 kg = 660000 g    (Conversion factor:  1 kg = 1000 g)

Putting values in above equation, we get:

[tex]\text{Mass percent of element}=\frac{47.5g}{660000g}\times 100=0.0072\%[/tex]

Hence, the mass percent of element in the ore is 0.0072 %

The answer is: mass of the snack mix can you consume and still be within the FDA limit is 491 grams.

ω(Na) = m(Na) ÷ m(salt).

ω(Na) = 39.33 g ÷ 100 g.

ω(Na) = 0.3933; mass percentage of sodium in salt.

m(salt) = 2.4 g ÷ 0.3933.

m(salt) = 6.10 g; mass of salt recommended for consumation.

Make proportion: 1.24 g : 100 g = 6.10 g ÷ m(mix).

m(mix) = 491.93 g; mass of the snack mix.