What is the density of a liquid with a mass of 6 grams, and a volume
of 3 milliliters?

Answer:

AuW₃

Explanation:

From the calculations, the change in energy is - 190 J.

From the first law of thermodynamics, the energy is neither created nor destroyed but is transformed from one form to another.

From the law;

U = q + w

U = internal energy

q = heat

w = work

Since work is done on the surroundings and the gas absorbs heat then;

U = 132 J - 322 J

U = - 190 J

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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.

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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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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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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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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The mole ratio would be 4:3

Mole ratio is the ratio of the stoichiometric number of moles of components of reactions.

From the equation: [tex]4NH_3 + 3O_2 --- > 2N_2 + 6H_2O[/tex]

The mole ratio of [tex]NH_3[/tex] to that of [tex]N_2[/tex] is the ratio of their coefficients. The coefficients are 4 and 3 respectively

Thus, the ratio is 4:3.

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Answer:

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

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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Answer:

AgNO3 + KI → AgI + KNO3

Explanation:

Yes, it's produced potassium nitrate

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:

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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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.

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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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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Answer:

1470 kJ (to 3 sf)

Explanation:

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

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]

From the equation of the graph, the activation energy of the reaction is -2.2 * 10⁴ J/mol.

Activation energy is the minimum required energy that reactant molecules must possess for a reaction to proceed towards formation of products.

The activation energy is determined from the slope of ln k against 1/T(K).

Given the equation, y = -2.2 * 10⁴ x + 45.0.

Comparing with the equation of a straight line, y = mx + c

The gradient, m = -2.2 * 10⁴

Therefore, the activation energy of the reaction is -2.2 * 10⁴ J/mol.

In conclusion, reactant molecules must break the activation energy barrier in order to form products.

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Answer:

3666.33 gm

Explanation:

O3  mole mass = 47.997

Each mole has   6.022 x 10^23   molecules

4.6 x 10^25 / 6.022 x 10^23     x    47.997 gm/mole  = 3666.33 gm

The hydrogen bonding atoms in melatonin are phospholipids and cholesterol.

This is the type of interaction which involves a hydrogen atom being covalently bound to a more electronegative "donor".

Melatonin controls sleep-wake cycle and has phospholipids and cholesterol which have high affinity for hydrogen bonding.

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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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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.

Answer:

carbon molecules share their 4 valence electrons through single, double, and triple bonds.

Explanation:

I think you meant ionic bonds not lonic bonds

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.

Additional:

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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)

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 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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The answers include:

This involves the formation of a new products from substances. In this scenario, a rising bread contains alcohol which evaporates.

Photographs also fall under this category and is therefore an irreversible chemical change.

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Answer:

The rate constant of the reaction at 125˚ is [tex]0.3115 \ \text{sec}^{-1}[/tex].

Explanation:

The Arrhenius equation is a simple equation that describes the dependent relationship between temperature and the rate constant of a chemical reaction. The Arrhenius equation is written mathematically as

                                                  [tex]k \ = \ Ae^{\displaystyle\frac{-E_{a}}{RT}}[/tex]

                                               [tex]\ln k \ = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT}[/tex]

where [tex]k[/tex] is the rate constant, [tex]E_{a}[/tex] represents the activation energy of the chemical reaction, [tex]R[/tex] is the gas constant, [tex]T[/tex] is the temperature, and [tex]A[/tex] is the frequency factor.

The frequency factor, [tex]A[/tex], is a constant that is derived experimentally and numerically that describes the frequency of molecular collisions and their orientation which varies slightly with temperature but this can be assumed to be constant across a small range of temperatures.

Consider that the rate constant be [tex]k_{1}[/tex] at an initial temperature [tex]T_{1}[/tex] and the rate constant [tex]k_{2}[/tex] at a final temperature [tex]T_{2}[/tex], thus

                         [tex]\ln k_{2} \ - \ \ln k_{1} = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT_{2}} \ - \ \left(\ln A \ - \ \displaystyle\frac{E_{a}}{RT_{1}}\right) \\ \\ \\ \rule{0.62cm}{0cm} \ln \left(\displaystyle\frac{k_{2}}{k_{1}}\right) \ = \ \displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)[/tex]

                                         [tex]\rule{1.62cm}{0cm} \displaystyle\frac{k_{2}}{k_{1}} \ = \ e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)} \\ \\ \\ \rule{1.62cm}{0cm} k_{2} \ = \ k_{1}e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)}[/tex]

Given that [tex]E_{a} \ = \ 26.5 \ \ \text{kJ/mol}[/tex], [tex]R \ = \ 8.3145 \ \ \text{J mol}^{-1} \ \text{K}^{-1}[/tex], [tex]T_{1} \ = \ \left(40 \ + \ 273\right) \ K[/tex], [tex]T_{2} \ = \ \left(125 \ + \ 273\right) \ K[/tex], and [tex]k_{1} \ = \ 0.0354 \ \ \text{sec}^{-1}[/tex], therefore,

           [tex]k_{2} \ = \ \left(0.0354 \ \ \text{sec}^{-1}\right)e^{\displaystyle\frac{26500 \ \text{J mol}^{-1}}{8.3145 \ \text{J mol}^{-1} \ \text{K}^{-1}}\left(\displaystyle\frac{1}{313 \ \text{K}} \ - \ \displaystyle\frac{1}{398 \ \text{K}} \right)} \\ \\ \\ k_{2} \ = \ 0.3115 \ \ \text{sec}^{-1}[/tex]