Given that the molar mass of the Red 40 dye is equal to 496.42 g/mol, how many grams of Red 40 food dye were in the 0.235 L solution of drink mix

Answer:

Molecules are moving the fastest and are the farthest apart in the gaseous phase.

Explanation:

Answer:

[tex]M=1.02M[/tex]

Explanation:

Hello.

In this case, since the molarity of a solution is computed as the moles of solute over volume of solution:

[tex]M=\frac{n_{solute}}{V}[/tex]

Since the volume of the solution is 257 mL or also 0.257 L, we compute the moles of solute, which is magnesium nitrate (molar mass = 148.3 g/mol), as shown below:

[tex]n_{solute}=39.0gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.3 gMg(NO_3)_2}=0.263molMg(NO_3)_2[/tex]

Thus, the required molarity turns out:

[tex]M=\frac{0.263molMg(NO_3)_2 }{0.257L}\\ \\M=1.02M[/tex]

Best regards.

Answer:

heterogeneous mixture

Explanation:

I'm not sure what halo halo or buko is but with mixed nuts you can see the mixture elements so that would be heterogeneous.

If you can't see the individual pieces in a mixture like ex: milk. that is a homogenous mixture.

Answer:

See explanation

Explanation:

a) Look at the electronic configuration of both species;

Cu^2+- [Ar]  3d9 4s0

Co^2+- [Ar]  3d7 4s0

We  can now see that a Jahn Teller effect occurs in [CuCl4]2– due to its d9 configuration. This  Jahn-Teller distortion causes the symmetry of  [CuCl4]2–  to change from tetrahedral  T

d  to trigonal pyramidal  C3v.

b) This is a consequence of Jahn-Teller effect in [CoF6]3– which makes the Co-F bonds lie along the z-axis and become elongated in comparison  to the remaining four Co-F bonds

Answer:

14.9%m/m

Explanation:

The percent m/m concentration of a solution (%m/m) indicates the grams of solute there are in 100 g of solution. It can be calculated as follows:

[tex]%m/m= \frac{mass of solute (g)}{mass of solution(g)} x 100[/tex]%m/m= mass of solute(g)/mass of solution(g) x 100

As the solution is composed by solute + solvent, the mathematical expression can be transformed in:

[tex]%m/m= \frac{mass of solute (g)}{mass of solution(g)} x 100[/tex]%m/m= mass of solute(g)/(mass of solute + mass of solvent (g) )x 100

         = 24.34 g[tex]%m/m= \frac{mass of solute (g)}{mass of solution(g)} x 100[/tex]/(24.34 g + 138.87 g) x 100

         = 24.34 g/(163.21 g) x 100

         = 14.9%

[tex]%m/m= \frac{mass of solute (g)}{mass of solution} x 100 = \frac{mass of solute (g)}{mass of solute (g) + mass of solvent (g)}[/tex]

Answer:

I got e but you might want to wait for someone else to answer it just incase...

Answer:

Magma cooling = metamorphic rock formation

Explanation:

Did you mean formation not folio?

Answer:

267.498 J is required to expand the volume of a pump from 0.0 L to 2.2 L against an external pressure of 1.2 atm

Explanation:

The volume change work of a gas is the work necessary for the gas to go from an initial volume Vi to a final volume Vf. If the volume decreases, the gas will have been compressed and it is a compression work; whereas if the volume increases, the gas will have expanded and it is expansion work.

The work is calculated by:

W= - P*ΔV= -P*(Vfinal - Vinitial)

In this case:

Replacing:

W= -121590 [tex]\frac{N}{m^{2} }[/tex]*0.0022 m³= - 267.498 N*m= -267.498 J

267.498 J is required to expand the volume of a pump from 0.0 L to 2.2 L against an external pressure of 1.2 atm

The work require to expand the volume of the given pump is 267.5 J.

The given parameters;

The work require to expand the volume of the given pump is calculated as follows;

W = PΔV

W = 1.2 atm (2.2 L - 0 L)

W = 2.64 atm.L

[tex]W = 2.64 \ atm.L \times \frac{101325 \ Pa}{1 \ atm} \times \frac{0.001 \ m^3}{1 \ L} \\\\W = 267.5 \ J[/tex]

Thus, the work require to expand the volume of the given pump is 267.5 J.

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Answer:

rip you

Explanation:

Answer:

0.44 M

Explanation:

Answer:

Both empirical and molecular formulas are NH₃.

Explanation:

Hello.

In this case, since we know the percent compositions of both nitrogen and hydrogen, in order find out the molecular formula and the empirical formula of that compound, we can first assume we have 82.27 g of nitrogen and 17.76 g of hydrogen as a first approximation to compute the present moles:

[tex]n_N=82.27g/(14.01g/mol)=5.872mol\\\\n_H=17.76g/(1.01g/mol)=17.58mol[/tex]

Next, we compute the subscripts in the empirical formula first, by diving the moles of each element by the moles of the element with fewest moles:

[tex]N:\frac{5.872}{5.872}=1\\\\H:\frac{17.58}{5.872}=3[/tex]

It means that the empirical formula is NH₃ and its molar mass is about 17.04 g/mol which is about the same the reported molar mass slightly more than 17 g, it means that the molecular formula is also NH₃.

Best regards!

Answer:

2 e⁻ + 4 H⁺(aq) + MnO₂(s) → Mn²⁺(aq) + 2 H₂O(l)

Explanation:

Let's write the half-reaction for the reduction of solid manganese dioxide MnO₂ to manganese ion Mn²⁺ in acidic aqueous solution.

MnO₂(s) → Mn²⁺(aq)

To balance the oxygen atoms, we will add 2 moles of water on the right side. To compensate for the hydrogen atoms added, we will add 4 hydrogen ions on the left side.

4 H⁺(aq) + MnO₂(s) → Mn²⁺(aq) + 2 H₂O(l)

Finally, we will balance the charges by adding 2 electrons on the left side.

2 e⁻ + 4 H⁺(aq) + MnO₂(s) → Mn²⁺(aq) + 2 H₂O(l)

Answer:

a

Explanation:

Answer:

B) Group 1 and 17

Explanation:

alkali metals and the halogens are the most reactive because of valence electrons

Answer:

Find the molar mass of all the elements in the compound in grams per mole.

Find the molecular mass of the entire compound.

Divide the component's molar mass by the entire molecular mass.

You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.

Explanation:

Mass spectrometry is used to measure the mass of atoms and to determine the percent composition of compounds.

There have been qualitative and numerical implications for mass spectroscopy. These include distinguishing unrecognized compounds, determining out a molecule's elemental isotopic makeup, and figuring out a compound's structure by analyzing at how it fragmented.

In order to determine the mass-to-charge ratio (m/z) of one or more molecules present in a sample, mass spectrometry is a proven to be a valuable instrument. The precise molecular weight of the sample's individual components may typically be estimated using these measures and also.

Thus, mass spectrometry is used to measure the mass of atoms and to determine the percent composition of compounds.

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Answer:

its polar

Explanation:

Answer:

(a) Map view and (b) cross-section view of location results for lower and upper groups overlapping with faults. Green dashed lines, magenta dots and blue dots refer to preexisting faults, lower events and upper events, respectively. (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)

(a) Map view and (b) cross-section view of location results for lower and upper groups overlapping with faults. Green dashed lines, magenta dots and blue dots refer to preexisting faults, lower events and upper events, respectively. (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)

Answer:

1011010101010100101010101101010101010010101010110101010101001010101011010101010100101010101101010101010010101010110101010101001010101011010101010100101010101101010101010010101010110101010101001010101011010101010100101010101101010101010010101010110101010101001010101011010101010100101010101101010101010010101010110101010101001010101011010101010100101010101101010101010010101010110101010101001010101011010101010100101010101101010101010010101010110101010101001010101011010101010100101010101101010101010010101010110101010101001010101011010101010100101010101101010101010010101010110101010101001010101011010101010100101010, You're welcome

Explanation:

Answer:

1.98 M

Explanation:

Step 1: Write the balanced reaction

2 NH₃ + 3 BrO⁻⟶ N₂ + 3 Br⁻ + 3 H₂O

Step 2: Calculate the moles corresponding to 0.325 g of ammonia

The molar mass of NH₃ is 17.03 g/mol.

0.325 g × (1 mol/17.03 g) = 0.0191 mol

Step 3: Calculate the reacting moles of hypobromite

The molar ratio of NH₃ to BrO⁻ is 2:3. The reacting moles of hypobromite are 3/2 × 0.0191 mol = 0.0287 mol

Step 3: Calculate the molar concentration of the hypobromite solution

M = 0.0287 mol / 0.0145 L = 1.98 M

The concentration of the hypobromite solution (BrO⁻) needed for the reaction is 1.97 M

We'll begin by calculating the number of mole in 0.325 g of NH₃. This can be obtained as follow:

Mass of NH₃ = 0.325 g

Molar mass of NH₃ = 17.031 g/mol

Mole = mass / molar mass

Mole of NH₃ = 0.325 / 17.031

Next, we shall determine the number of mole of BrO⁻ needed to react with 0.01908 mole of NH₃. This can be obtained as follow:

2NH₃ + 3BrO⁻ —> N₂ + 3Br⁻ + 3H₂O

From the balanced equation above,

2 moles of NH₃ reacted with 3 moles of BrO⁻.

Therefore,

0.01908 mole of NH₃ will react with = [tex]\frac{0.01908 * 3 }{2}\\\\[/tex] = 0.02862 mole of BrO⁻.

Finally, we shall determine the concentration of the BrO⁻ solution.

Mole of BrO⁻ = 0.02862 mole

Volume of BrO⁻ solution = 14.5 mL = 14.5 / 1000 = 0.0145 L

Concentration = mole / Volume

Concentration of BrO⁻ = 0.02862 / 0.0145

Therefore, the concentration of the hypobromite solution (BrO⁻) is 1.97 M

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Answer:

Paper, some plastics, frosted glass, etc.

Explanation:

Answer:

Positive charged

Explanation:

Answer:

False

Explanation:

Answer:

-183.15 °C

Explanation:

Given data:

Mass of F₂ = 5.00 g

Pressure = 119.96 KPa (119.96/101 = 1.2 atm)

Volume = 750.0 mL ( 750.0/1000 = 0.75L)

Temperature = ?

Solution:

First of all we will calculate the number of moles.

Number of moles = mass/molar mass

Number of moles = 5.00 g/ 38g/mol

Number of moles = 0.13 mol

Now we will calculate the temperature:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1.2 atm × 0.75 L = 0.13 mol ×0.0821 atm.L/ mol.K  × T

0.9 atm.L = 0.01 atm.L/ K  × T

T = 0.9 atm.L /0.01 atm.L/ K

T = 90 K

Kelvin to degree celsius:

90K -273.15 = -183.15 °C

Here is the full question:

A chemistry graduate student was given 500 mL of a 1.60 M hydrocyanic acid (HCN) solution. Hydrocyanic acid is a weak acid with [tex]K_a = 4.9 \times 10^{-10}[/tex]. What mass of KCN should the student dissolve in the HCN solution to turn it into a buffer with pH  = 9.16?

Answer:

36.85 g

Explanation:

From the given information above:

Recall that  number of moles of HCN = mass of HCN / molar mass of HCN= mass of HCN / molar mass of HCN

Therefore;

the number of moles of HCN =  [tex]\dfrac{500 \times 1.60}{1000}[/tex]

the number of moles of HCN = [tex]\dfrac{800}{1000}[/tex]

the number of moles of HCN =  0.8

The acid dissociation constant [tex]K_a[/tex] = [tex]4.9 \times 10^{-10}[/tex]

Thus the pKa = [tex]- log (4.9 \times 10^{-10})[/tex]

pKa = 9.31

However, the pH is given as:

[tex]pH = pKa + log \bigg ( \dfrac{KCN}{HCN} \bigg )[/tex]

[tex]9,16 =9.31 + log \bigg ( \dfrac{KCN}{0.8} \bigg )[/tex]

[tex]9,16 -9.31 =log \bigg ( \dfrac{KCN}{0.8} \bigg )[/tex]

[tex]-0.15 =log \bigg ( \dfrac{KCN}{0.8} \bigg )[/tex]

KCN = 0.566 mol

The standard molar mass of KCN = 65.11 g/mol

Thus, the mass of KCN = number of moles of KCN × molar mass of KCN

the mass of KCN = 0.566 × 65.11

the mass of KCN that the student needs to dissolve in HCN is =  36.85 g

Explanation:

Then there is no net charge separation. a BH3 b CHCl3 c C2H2 d NH3. Optionally the ... Oxygen O2 c. ... BH3 2. Com Bh3 Polar Or Nonpolar In the case of non polar molecules dispersion forces or London forces are present between them.

The polar molecules is NO.

A polar molecule is one that has one end that is slightly positive and the other end that is slightly negative. Ionic or polar covalent bonds can form in polar molecule. A dipole is a molecule that has two poles. The dipole moment is the outcome of measuring the amount of polarity in a molecule.

NO molecule, In the periodic table, nitrogen is to the right of oxygen. Nitrogen has a lower electronegative potential than oxygen. All N-O bonds are polar, with the oxygen atom having a higher electron density. The uneven distribution of electrons in the NO molecule is quite minor. When a molecule is non-polar, the electrons are shared evenly, resulting in a non-polar bond, or the polar bonds are symmetric. So, oxygen, nitrogen and boron trihydride molecules are non-polar molecule.

Hence the correct option is D.

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Answer: The freezing point of the solution will be [tex]4.3^0C[/tex]

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=T_f^0-T_f=(5.5-T_f)^0C[/tex] = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

[tex]K_f[/tex] = freezing point constant = [tex]4.90^0C/m[/tex]

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (benzene)= 800 g = 0.8 kg

Molar mass of solute ( methyl salicylate) = 152.1 g/mol

Mass of solute ( methyl salicylate) = 30 g

[tex](5.5-T_f)^0=1\times 4.90^0C/m\times \frac{30g}{152.1 g/mol\times 0.8kg}[/tex]

[tex](5.5-T_f)0C=1.2[/tex]

[tex]T_f=4.3^0[/tex]

The freezing point of the solution will be [tex]4.3^0C[/tex]

Answer:

21.95 g of AgCl.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

AlCl3(aq) + 3AgNO3(aq) —> 3AgCl(s) + Al(NO3)3(aq)

Next, we shall determine the masses of AlCl3 and AgNO3 that reacted and the mass of AgCl produced from the balanced equation. This is illustrated below:

Molar mass of AlCl3 = 27 + (3×35.5) = 133.5 g/mol

Mass of AlCl3 from the balanced equation = 1 × 133.5 = 133.5 g

Molar mass of AgNO3 = 108 + 14 + (3×16) = 170 g/mol

Mass of AgNO3 from the balanced equation = 3 × 170 = 510 g

Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Mass of AgCl from the balanced equation = 3 × 143.5 = 430.5 g

From the balanced equation above,

133.5 g of AlCl3 reacted with 510 g of AgNO3 to produce 430.5 g of AgCl.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

133.5 g of AlCl3 reacted with 510 g of AgNO3.

Therefore,

22 g of AlCl3 will react with =

(22 × 510)/133.5 = 84.04 g of AgNO3.

Thus, we can see that it will require a higher mass (i.e 84.04 g) of AgNO3 than what was given (i.e 26 g) to react completely with 22 g of AlCl3. Therefore, AgNO3 is the limiting reactant and AlCl3 is the excess reactant.

Finally, we shall determine the mass of the solid product (AgCl) produced from the reaction.

In this case the limiting reactant will be used because, it will produce the maximum yield of the reaction since all of it is used up in the reaction.

The limiting reactant is AgNO3 and the mass of AgCl produced can be obtained as follow:

From the balanced equation above,

510 g of AgNO3 reacted to produce 430.5 g of AgCl.

Therefore, 26 g of AgNO3 will react to produce = (26 × 430.5)/510 = 21.95 g of AgCl.

Thus, the mass of the solid product (AgCl) produced is 21.95 g

Answer:

7.09 cm³

Explanation:

From the question given above, the following data were obtained:

Density of Germanium = 5.32 g/cm³

Mass of Germanium = 37.72 g

Volume of Germanium =?

The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Mathematically it is expressed as:

Density = mass / volume

Thus, with the above formula, we can obtain the volume of Germanium as shown below:

Density of Germanium = 5.32 g/cm³

Mass of Germanium = 37.72 g

Volume of Germanium =?

Density = mass /volume

5.32 = 37.72 / volume

Cross multiply

5.32 × Volume = 37.72

Divide both side by 5.32

Volume = 37.72 / 5.32

Volume of Germanium = 7.09 cm³

Answer:

λ [Wavelength] = 3.04357825 m [meters]

Explanation:

Given a frequency, to find the wavelength, divide the speed of light by the frequency. Wavelength = Speed Of Light In Seconds / Frequency .

Wavelength in meters is: λ = 299,792,458 m/s / f Hz .

Then substitute in the frequency to find the wavelength: λ = 299,792,458 m/s / f Hz → λ = 299,792,458 m/s / 9.85 × 10^7 Hz → λ = 2.99792458 × 10^8 m/s / 9.85 × 10^7 → λ = 2.99792458 × 10^8 m / 9.85 × 10^7 → λ = 29.9792458 / 9.85 m → λ = 3.04357825 m

Answer:

Some distortion exists somewhere in the map.

Explanation:

Any map that shows the entire world will have some distortion somewhere in the map. This is because we are transforming a spherical body to a flat rectangular dimensional surface.