What is the mass of 8 moles of sodium chloride?
468 g
4.68 9
46.8 g
4.689 mol
Please answer
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Explanation:

the investigation lasts for 7 days !

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The sample of white powder contains 47.1% K2CO3 and 0.39% Na2CO3.

Molar mass of sodium carbonate = 106 g/mol

Molar mass of potassium carbonate = 138 g/mol

Number of moles of HNO3 = 10/1000 L × 0.8903 M = 0.008903 moles

Mass of HNO3 =  0.008903 moles × 63 g/mol = 0.56 g

Mass of sample added = 1.091 g

Mass of sample left over = 0.573 g

Mass of sample reacted = 1.091 g - 0.573 g = 0.518 g

The reacted sample contains xg of Na2CO3 and (0.518 - x) g K2CO3.

Na2CO3 + 2HNO3 --> 2NaNO3 + CO2 + H2O

106g of Na2CO3 reacts with 126g of HNO3

x g of Na2CO3 reacts with (126 × x/106)g of HNO3

K2CO3 + 2HNO3 --> 2KNO3 + CO2 + H2O

138 g of K2CO3 reacts with 126 g of HNO3

(0.518 - x) g of K2CO3 reacts with [(0.518 - x) × 126/138] g

Total mass of HNO3 used;

1.19x + 0.47 + 0.91x = 0.56

2.1x + 0.47 = 0.56

2.1x = 0.56 - 0.47

2.1x = 0.09

x =  0.09/2.1

x = 0.0043 g

Mass of K2CO3 = (0.518 - x) g = 0.518 - 0.0043 = 0.5137 g

Mass percent of K2CO3  = 0.5137 g/ 1.091 g × 100/1 = 47.1%

Mass percent of Na2CO3 = 0.0043/1.091 g × 100/1 = 0.39%

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Answer:

-85 °C

Explanation:

O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.

H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter

molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.

Hey there!

It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.

In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:

[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]

Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:

[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]

Thus, the density of the calcite sample will be:

[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]

This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL

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Regards!

Answer:

The answer is option A

Explanation:

This is because the measure from crest to crest is also known as the wavelength, and the color of a wave always depends on its wavelength, if you see the electromagnetic spectrum

Hope this helps!

Explanation:

A=Sun

B=Earth

C=Neptune.

hope my answer will definitely help you out.

by bouncing it

Explanation:

when you bouce a ball it hits the ground because gravity pulls it down

Answer:

Explanation:

we will do ratio method

Aluminum chloride :  Aluminum

2                               :      2

0.48                          :      x

(cross multiply)

0.48 x 2 / 2 = 0.48 moles of aluminum  

mass = 1 mole of aluminum chloride x moles

mass = 133.33 x 0.48

mass = 63.9984g (round off) = 64g

I hope this helps.

Answer:

from the words below underline six example of rhetorical patterns

Answer:

read the story

Explanation:

read it and then you will find the answer

If a student notices that  one-quarter of the moon is visible, a lunar eclipse has taken place.

Eclipse is said to occur when one of the heavenly bodies casts a shadow on another heavenly body causing it to be partly visible or completely invisible. When the earth comes between the sun and the moon, a lunar eclipse is said to occur.

When a lunar eclipse occurs, the moon may be partly visible or completely invisible. Therefore, if a student notices that  one-quarter of the moon is visible, a lunar eclipse has taken place.

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Answer:

bladder or bowel diseases, or arthritis. These might be your answer.

Based on our knowledge of strong and weak acids, we can confirm that the Ka value for acetic acid will be relatively low since it is a weak acid.

Acids can be strong or weak. This is determined by its tendency to break apart into ions or stay together to form molecules. Although somewhat counter-intuitive, strong acids are those that are most likely to break apart and therefore contain a high number of ions within their solutions.

Weak acids, on the other hand, are those that tend to stay together in the form of molecules and therefore possess very low ion counts in their solutions. The acid dissociation constant, Kₐ, is used to measure whether an acid is weak or strong and how much so. In the case of Acetic acid, the ka measurement will offer a low value, indicating a weak acid.

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Explanation:

Oils are made up of long hydrocarbon chains which are non polar. Since water is a polar solvent oil doesn't dissolve in water well.( Polar solutes dissolve in polar solvents and non polar solutes dissolve in non polar solvents) The hydrocarbon chains are hydrophobic.

Answer:

The number by which we divide is called the divisor. The result obtained is called the quotient. The number left over is called the remainder.

Explanation:

Vectors may be plasmids. An example of one of several useful features of commercially prepared cloning vectors is MULTIPLE CLONING SITES.

The pUC18 vector is a widely used standardized cloning vector for replication in Escherichia coli.

A multiple cloning site can be defined as a short DNA fragment observed in genetically engineered plasmids.

These DNA fragments (multiple cloning sites) contain twenty (20) or more sites where restriction enzymes can cut in order to generate recombinant DNA molecules.

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Explanation:

since 5g saturates 12.5g of water at 10°c

so......x would saturate 50g of water at 0°c

then you can cross multiply

Answer:

[tex]Molarity\,\,of\,\,the\,\,solution\,\,is\,\,S=0.039M[/tex]

Explanation:

[tex]W=2.18 g\\M=322g\\V=175mL=0.175L\\\\S=\frac{W}{MV} \\=>S=\frac{2.18}{322*0.175} \\So,S=0.039M[/tex]

Hope you have understood this.....            

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From the data obtained from the tests, the actual boiling point lies between 34°C and 40°C.

The boiling point of a substance is the temperature at which the pressure of the substance becomes equal to atmospheric pressure. Pure substances have a sharp boiling point while impure substances boil over a temperature range.

In this case, the boiling point of the substance after 10 tests is obtained as 37+/−3°C. This implies that the actual boiling point lies between 34°C and 40°C.

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Explanation:

Electroplated jeweleries are in demand because firstly they are as shiny and attractive as real jeweleries, they are also light weighted and cost effective.

Three resonance structures  can be drawn for the allyl cation while two resonance structures can be drawn for the amidate ion.

Sometimes, we cannot fully describe the bonding in a chemical specie using a single chemical structure. In such cases, we have to use a number of structures which cooperatively represent the actual bonding in the molecule. These structures are called resonance or canonical structures.

The resonance structures of the allyl cation and the amidate ion are shown in the images attached to this answer. These structures show the different bonding extremes in these organic ions.

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Answer:

composition of an element

This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.

The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.

However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.

As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:

[tex]y=-3.5 x + 25\\\\y=-0.52 x + 2[/tex]

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

[tex]-3.5 x + 25=-0.52 x + 2\\\\-3.5 x+0.52 x =2-25\\\\x=\frac{-23}{-2.98}=7.72[/tex]

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

[tex]y=-3.5 (7.72) + 25\\\\y = 1.84[/tex]

This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.

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Answer:

the answer would be B

Explanation:

The power from the charger use electricity and the electricity transforms to energy that can be moved

Answer:

New volume = 6.0L

Explanation:

Make a list of variables given in the question, and what we want to find:

Initial volume (L) = Vi = 3.1

Initial temperature (K) = Ti = 193

New temperature (K) = Tn = 374

New volume (L) = Vn

We have volumes and temperatures given and we're talking about gas, the equation including both these variables that should come to mind is the ideal gas equation:

PV = nRT

P = pressure (Pa)

n = number of moles

R = gas constant = 8.31

Now, we construct an equation we can solve to get the new volume:

P(Vn) = nR(Tn)

And insert the values we know:

P(Vn) = n(8.31)(374)

P(Vn) = n·3107.94

The equation contains 3 unknown variables;

We should first try to see if we can eliminate the 2 we are not interested in, namely P and n;

What we should recognise is that the pressure and number of moles will not change upon warming the balloon;

Firstly, heating the gas inside the balloon doesn't add anything to the balloon, i.e. doesn't increase the moles of gas, it simply raises the energy of the gas particles already within the balloon, so n will not change;

Secondly, I think there is a mistake in the question, it should read "assume constant pressure" in the brackets, since the temperature does change, which we are told;

Also, we can assume normal atmospheric pressure inside and outside the balloon as would be the case ordinarily;

What we want to do with this information is rearrange the equation we constructed to have these 2 constant or unchanging variables on one side and everything else on the other, so:

[tex]\frac{P}{n} = \frac{3107.94}{V_{n} }[/tex]

Next, we construct an equation for the balloon before warming:

P(Vi) = nR(Ti)

P(3.1) = n(8.31)(193)

P·3.1 = n·1603.83

Once again, rearrange to get P and n on one side of the equation and everything else on the other:

[tex]\frac{P}{n} = \frac{1603.83}{3.1} \\\\ \frac{P}{n} = 517.364516[/tex]

Now, we have two equations for P/n, we can eliminate P/n since both these values remain the same before and after warming the balloon as previously established:

[tex]\frac{3107.94}{V_{n} } = \frac{P}{n} = 517.364516 \\\\ \frac{3107.94}{V_{n} } = 517.364516 \\\\[/tex]

Now, rearrange and solve for Vn:

[tex]\frac{3107.94}{V_{n} } = 517.364516 \\\\ V_{n} = \frac{3107.94}{517.364516} \\\\ V_{n} = 6.00725...[/tex]

Vn = 6.0L

Answer:

A mixture of chlorine gas (Cl2) and fluorine gas (F2):

✔ London dispersion forces

Explanation:

Answer:

ask you subject teacher

Explanation:

Answer:

Michal Faraday 19th century

Answer:

We can do an experiment with vinegar to see what happens to the calcium in an eggshell when it is exposed to an acid.