4. It is desired to prepare 0.5 L of a 0.1 M solution of NaCl
from a 1 M stock solution. How many milliliters of the
stock solution must be taken for the dilution?
5. Calculate the osmotic pressure of a 0.5 M sucrose
solution if the temperature is 298 K.

The equilibrium expression for the reaction; C(s) + O₂(g) ------->CO₂(g) is [CO2]/[C] [O2] option C.

The equilibrium expression shows how much of the reactants are converted into products. If the equilibrium constant is large and positive, most of the reactants have been converted into products.

Thus, the equilibrium expression for the reaction; C(s) + O₂(g) ------->CO₂(g) is [CO2]/[C] [O2] option C.

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The pH of a 0.260 M solution of ascorbic acid is 0.585. Details about pH can be found below.

The pH of a solution can be calculated using the following expression:

pH = - log {H+}

According to this question, ascorbic acid is a diprotic acid and posseses a concentration of 0.260M. The pH can be calculated as follows;

pH = - log {0.260}

pH = 0.585

Therefore, the pH of a 0.260 M solution of ascorbic acid is 0.585.

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The percentage composition of Nitrogen in the compound is 35.71%

Using the ideal gas equation, the mole of nitrogen can be obtained as follow

PV = nRT

n = PV / RT

n = (700 × 0.1) / ( 62.364 × 250)

n = 0.00448 mole

Mass = mole × molar mass

Mass of nitrogen = 0.00448 × 28

Mass of nitrogen = 0.125 g

Percentage = (mass of element / mass of compound) × 100

Percentage of nitrogen = (0.125 / 0.35) × 100

Percentage of nitrogen = 35.71%

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Thermal energy depends not only on the temperature of a substance but also on the amount of it.

It may be helpful to think of this situation as looking for the total amount of molecular movement. The pot of boiling water has a lot of movement but only a few molecules. The iceberg has far less movement between molecules but vastly more of them.

With the size alone, the slight movements between the molecules of the iceberg add up immensely and tower over that same measurement on the pot of boiling water.

The amount of 1,3-dimethyl urea produced would be 32,458 grams or 32.458 kg while that of carbon dioxide would be 16,214 grams of 16.214 kg

From the equation of the reaction, the mole ratio of methyl isocyanate with the products is 2:1 respectively.

Mole of 42,000 kg of methyl isocyanate = 42000/57 = 736.84 moles

Equivalent mole of 1,3-dimethyl urea = 736.84/2 =368.42 moles

Equivaent mole of carbon dioxide = 736.84/2 =368.42moles

Mass of 368.42 moles 1,3-dimethyl urea = 368.42 x 88.1 = 32,458 grams or 32.458 kg

Mass of 368.42 moles of carbon dioxide = 368.42 x 44.01 = 16,214 grams of 16.214 kg

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The experiment seeks to find out if the seeds take up oxygen as they germinate.

An experiment is carried out in order to  determine cause and effect relationships. The independent variable is the variable that is manipulated in the experiment.

a. The set up B is the control set up

b. The independent variable is the presence/ absence of the seeds while the dependent variable is the color of the solution in both containers.

c. The amount of oxygen in each container can be measured before and after.

d. The hypothesis is that the seeds take up oxygen as they germinate.

Hence, the experiment seeks to find out if the seeds take up oxygen as they germinate.

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116.6 x [tex]10^{ -3[/tex]} is the volume of methane contained in 2.50 L of a saturated solution at 25 ∘C that was extracted and placed under STP conditions.

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

The solubility of methane in water at 25 degrees Celcius is 1.3 x [tex]10^{ -3}[/tex] M. It implies   1.3 x [tex]10^{ -3}[/tex] moles of methane are dissolved in one litre of water.

The number of moles of methane in 4 L of water can be calculated as follows:

Moles of methane = [tex]\frac{1.3 . 10^{ -3} }{1L}[/tex] x[tex]\frac{4L}{1L}[/tex]

Moles of methane = 5.2 x [tex]10^{ -3}[/tex]

STP refers to standard temperature and pressure. Under STP conditions, the temperature of the substance is  0 degrees celcius and its pressure is 1 atm.

An ideal gas is an imaginary gas comprising of a large number of randomly moving particles and the motion between such articles is considered to be perfectly elastic. The ideal gas equation describes the relationship between pressure, volume, temperature and number of moles of a gas.

The expression for ideal gas equation is as follows:

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1 atm

V= ?

R= [tex]0.082057338 \;L \;atm \;K^{-1}mol^{-1}[/tex]

T=273K

n=?

Putting value in the given equation:

[tex]\frac{PV}{RT}=n[/tex]

5.2 x [tex]10^{ -3} = \frac{1 \;atm\; X \;V}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 273}[/tex]

V= 116.6 x [tex]10^{ -3}[/tex]

Hence, 116.6 x [tex]10^{ -3[/tex]} is the volume of methane contained in 2.50 L of a saturated solution at 25 ∘C was extracted and placed under STP conditions.

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Answer:

AuW₃

Explanation:

The empirical formula of the compound is SiCl₄.

mass of silicon = 28 g/mol

mass of chlorine = 35.5 g/mol

Si = 8.56/28   =  0.306

Cl = 43.2/35.5  = 1.216

Si = 0.306/0.306 = 1

Cl = 1.216/0.306 = 4

empirical formula = SiCl₄

Thus, the empirical formula of the compound is SiCl₄.

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The Ksp of the solution is 1.97 * 10^-9.

The pH is defined as the hydrogen ion concentration of the solution. Now we have the pH of the solution as 10.9.

We know that;

pOH = 14 - pH

pOH = 14 - 10.9

pOH = 3.1

[OH^-] = Antilog [-3.1]

[OH^-] =7.9 * 10^-4 M

Now

Ksp = [M^2+] [2OH^-]^2

Ksp = s * (2s)^2

Ksp = 4s^3

Now s = 7.9 * 10^-4 M

Ksp = 4(7.9 * 10^-4)^3

Ksp = 1.97 * 10^-9

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Answer:

1470 kJ (to 3 sf)

Explanation:

(802)(1.83) = 1470 kJ (to 3 sf)

The volume of 6.40 grams of O₂ gas at STP is 4.48L (option A). Details about volume can be found below.

The volume of a gas can be calculated using the following formula:

p = m/v

Where;

According to this question, the mass of O₂ gas at STP is 6.40 grams. The density of the gas at STP is 1.43 g/L.

1.43g/L = 6.4g/V

Volume of O2 = 6.4 ÷ 1.43 = 4.48L

Therefore, the volume of 6.40 grams of O₂ gas at STP is 4.48L.

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Answer:

This is right!!! the answer is A 4.49

Explanation:

Boils

Celsius is based on the freezing and boiling points of water.

Water freezes below 0 degrees Celsius (32 degrees Fahrenheit), and it boils above 100 degrees Celsius (212 degrees Fahrenheit).

Answer:

Ice cubes do not melt faster in cold water than in hot water. Melting of ice cubes requires heat. What better source of heat than the hot water the ice cubes are in.

Answer:

AgNO3 + KI → AgI + KNO3

Explanation:

Yes, it's produced potassium nitrate

Answer:

320

Explanation:

Because the question is asking for 2 significant figures, you want to try to get rid of all of the numbers to the right of the 1. That way, the only numbers that will be significant are the numbers in the position of the 3 and the 1.

Before you can make the rest of the numbers equal 0, you need to correctly round the number in the last significant position (the number 1).

If the number to the right of the 1 is from 0-4, the number stays the same. If the number is from 5-9, the 1 needs to be increased by a unit.

Since the number to the right is an 8, you need to change the 1 to a 2. Now, you can make all of the other numbers a 0. There should be no decimal place because it would mean that the zero to the left of it is significant.

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \:mass = 11.42 \:\: grams [/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

[tex]\qquad \tt \rightarrow \: Q = ms\Delta T[/tex]

[tex] \large\textsf{Find m : } [/tex]

[tex]\qquad \tt \rightarrow \: 501 = m \sdot(0.45) \sdot(97.5)[/tex]

[tex]\qquad \tt \rightarrow \: 501 = m \sdot43.875[/tex]

[tex]\qquad \tt \rightarrow \: m = \dfrac{501}{43.875} [/tex]

[tex]\qquad \tt \rightarrow \: m \approx11.42 \: \: g[/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

The electromagnetic spectrum of an element is not limited only to the visible region of the spectrum.

An electromagnetic spectrum is a range of frequencies and wavelengths into which an electromagnetic wave is distributed into.

In Science, the electromagnetic spectrum consist of the following types of energy from highest to lowest frequency and shortest to longest wavelength:

In this context, we can infer and logically deduce that the electromagnetic spectrum of an element is not limited only to the visible region of the spectrum.

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Taking into account the reaction stoichiometry, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

In first place, the balanced reaction is:

3 Cu+ 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The following rule of three can be applied: if by reaction stoichiometry 504 grams of HNO₃ form 562.65 grams of Cu(NO₃)₂, 4.69 grams of HNO₃ form how much mass of Cu(NO₃)₂?

[tex]mass of Cu(NO_{3} )_{2} =\frac{4.69 grams of HNO_{3} x562.65 grams of Cu(NO_{3} )_{2} }{504 grams of HNO_{3} }[/tex]

mass of Cu(NO₃)₂=  5.2634 grams

Then, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

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21.1 grams of calcium chloride will be produced when 27.0 g of calcium carbonate is combined with 14.0 g of hydrochloric acid.

The reactant that is consumed first in a chemical reaction is the limiting reagent or limiting reactant because it stops any more reactions from taking place. The limiting reagent controls how much product is produced during the reaction.

CaCO3(s) + 2HCl(aq) ==> CaCl2(aq) + H2O(l) + CO2(g)  

[tex]mols CaCO3 =[/tex] [tex]27 *\frac{1 mol}{100g}[/tex] = 0.27 mol

[tex]mols HCl = 15 gx \frac{1 mol}{36.5 g}[/tex] = 0.38 mol

Limiting reactant = HCl ( it takes 2x the mols of HCl compared to mols CaCO3 as per balanced equation)

mass of CaCl2 = [tex]0.411 mols HCl x\frac{ 1 mol CaCl2}{ 2 mols HCl } x 111.0 g/mol[/tex]

mass of CaCl2 = 21.1 g

Hence, 21.1 g of calcium chloride will be produced when 27.0 g of calcium carbonate is combined with 14.0 g of hydrochloric acid.

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When an electron goes from a higher shell to a lower shell then it loses energy.

A photon is a particle of light defined as a discrete bundle (or quantum) of electromagnetic (or light) energy.

When an electron goes from a higher shell to a lower shell then it loses energy.

So, when an electron moved from shell n = 2 to shell n = 1 then a photon of energy is released.

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Taking into account the Charles's law, the gas occupies a volume of 0.0182 m³ when the temperature is raised to 95.0 °C. ​

Charles's law shows the relationship between the volume and temperature of a gas sample at constant pressure.

This law states that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles' law is a law that says that when the amount of gas and pressure remain constant, the ratio between volume and temperature will always have the same value:

[tex]\frac{V}{T}= k[/tex]

Considering an initial state 1 and an initial state 2:

[tex]\frac{V1}{T1}= \frac{V2}{T2}[/tex]

In  this case, you know:

Replacing in Charles's Law:

[tex]\frac{0.0141 m^{3} }{284.5 K}=\frac{V2}{368 K}[/tex]

Solving:

[tex]\frac{0.0141 m^{3} }{284.5 K}x368 K=V2[/tex]

0.0182 m³= V2

Finally, the gas occupies a volume of 0.0182 m³ when the temperature is raised to 95.0 °C. ​

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Answer:

cannot remain at rest under the action of any shear force.

From the coefficients of the equation, we know that for every 2 moles of Al consumed, 6 moles of HCl are consumed.

This means 6(2/2)=6 moles of HCl are needed.

Since there are [tex]6.022 \times 10^{23}[/tex] molecules in 1 mole of a substance, there are :

[tex]6(6.022 \times 10^{23}}=\boxed{4 \times 10^{24} \text{ (to 1 sf)}}[/tex]

1. The volume of the cube (cm³) is 0.599 cm³

2. The density of the copper to two decimal place is 8.91 g/cm³

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

V = L³

V = 0.843³

V = 0.599 cm³

Density = mass / volume

Density = 5.34 / 0.599

Density = 8.91 g/cm³

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Answer:

1. H₂SO₄

Explanation:

A balanced equation tells us how each reactant and product is related to each other in terms of moles.

The equation given has already been balanced since the number of atoms of each element is the same on both sides of the arrow. From the equation, we can see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Whichever reactant that we do not have enough of to make more products is known as the limiting reagent, since it limits the amount of products formed.

Mole ratio of H₂SO₄: NaOH= 1: 2

The amount of NaOH needed is twice the amount of H₂SO₄ in terms of moles.

Given that there are 0.40 moles of H₂SO₄, let's calculate the amount of NaOH needed.

Amount of NaOH needed

= 2(0.40)

= 0.80 moles

However, there are only 0.75 moles of H₂SO₄. Since 0.75 <0.80, H₂SO₄ is the limiting reactant.

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Answer:

See below

Explanation:

From the balanced equation you can see you need twice as many moles of NaOH as H2SO4....

   you have .4 moles of H2SO4   so you need .8 moles of NaOH

              but you only have .75 mole....so NaOH is the limiting factor (because you do not have enough)

Answer:

=> 0.10 moles

Explanation:

The question is asking us to find the moles of NaCl cid formed.

We have been provided with;

0.10 L of 1.0 M NaOH.

We know that mole ratio of the reaction is 1:1 and also the molarity of a solution is contained in 1 L or 1000 mL or 1000 cm³.

This means that;

1.0 M is contained in 1.0 L.

X mol is contained in 0.10 L

[tex] x \: mol \: = \frac{0.1 \times 1.0}{1.0} \\ = 0.10 \: mol[/tex]

= 0.10 moles of NaCl formed

Answer:

Na2CO3 + 2HC2H3O2 --> CO2 + H2O + 2NaC2H3O2 (note my corrections to your original post)

2Na^+ + CO3^-2 + 2HC2H3O2 ==> CO2 + H2O + 2Na^+ + 2C2H3O2^-

Now cancel the ions common to each side to arrive at the net ionic equation. To be a little more accurate, place an (aq) after each ion or solution.

ur correct answer

Explanation:

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The following types of bonds exists between the listed atoms;

A bond is polar covalent when the electronegativity difference between the bonding atoms is above that found in a nonpolar covalent bond.  Now we shall see the types of bonds between these atoms.

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