Based on your observations, relate the container from the simulation to a car airbag. What would happen to an inflatable airbag if air molecules were to be continuously added to it, similar to the container in the simulation?

The maximum amount of CaCO3 we can expect is 0.0180 mole x 100 g/mole = 1.80 g.

Given that:

There would be a precipitate of calcium carbonate from the reaction between sodium carbonate and calcium chloride.

The reaction is:

Na2CO3 (aq) + CaCl2(aq) → CaCO3 (s) + 2 NaCl (aq)

2.00 g CaCl2 x 1 mole/ 111 g CaCl2

= 0.0180 mole CaCl2

Mix it with 2.00 g Na2CO3 x 1 mole/ 106 g

= 0.0189 mole Na2CO3

The maximum amount of CaCO3 we can expect is 0.0180-mole x 100 g/mole = 1.80 g

Therefore, the 1.80 g is the theoretical (calculated) yield of CaCO3 in this example.

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Answer:

2 g/mL

Explanation:

Density can be calculated using the ratio:

Density = mass (g) / volume (mL)

Since you have been given both the mass and volume of the liquid, you can calculate the density.

Mass = 6 grams (g)

Volume = 3 milliliters (mL)

Density = mass / volume

Density = 6 g / 3 mL

Density = 2 g/mL

The scratched tin can with the iron will more rapidly corrode the iron than the tin.

The corrosion can be given as the process of the oxidation of the metal into the more stable metal oxide. The chemical oxidation of the metal is attained with the surrounding available oxygen or the water vapors.

The reactivity of the Iron for the oxidation is more as compared to the tin from the reactivity series. Therefore if both tin and iron are exposed to corrosion, iron will be more rapidly corroded.

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Answer:

65.4 mmHg

Explanation:

Using Boyle's Law,

(248)(1.05)=(pressure)(3.98)

pressure = 65.4 mmHg

Answer:

Noble gases are rare elements (gases) with extreme stability and a very low reactivity rate.

Some examples are neon, argon, helium etc.

Hope it helps!!

Answer:

Helium, neon, argon, krypton, xenon, and radon.

Explanation:

The chemical elements in group 18 of the periodic chart are the noble gases. They have the most valence electrons that their outer shell can support, making them the most stable. As a result of their inherent stability, they rarely interact with other components.

Other characteristics of the noble gases are that they all conduct electricity, fluoresce, are odorless and colorless, and are used in many conditions when a stable element is needed to maintain a safe and constant environment. These are the 6 noble gases:

Although the phrase "inert gases" was once used to describe the noble gases, it is not entirely true given that some of them do participate in chemical processes.

Chemical properties:

Under normal circumstances, the noble gases are nonflammable, tasteless, colorless, and odorless. Because it was earlier thought that they had a valence of zero, or that their atoms couldn't mix with those of other elements to create compounds, they were originally given the group 0 designation in the periodic table. However, it was eventually determined that some do in fact combine to create compounds, which led to the label's demise. Full valence electron shells are present in noble gases. The only electrons that often take part in chemical bonding are valence electrons, which are an atom's outermost electrons.  Due to their great stability, complete valence electron shell atoms have little to no inclination to gain or lose electrons and do not frequently establish chemical bonds. But unlike lesser noble gases like helium, larger noble gases like radon are bound together by electromagnetic force less tightly, making it simpler to remove outside electrons from hefty noble gases. The noble gases can be combined with the electron configuration notation to create the noble gas notation as a consequence of a complete shell.

To do this, the electron configuration is written starting with the nearest noble gas that comes before the element in issue, and then it is carried from there. For instance, the noble gas notation is [Ne] 3s2 3p3, whereas the electron notation for phosphorus is 1s2 2s2 2p6 3s2 3p3. In addition to being shorter than writing down the whole notation of atomic orbitals, this more condensed notation makes it simpler to identify elements.

Of all the noble gas compounds that have been created, xenon compounds are the most prevalent. [33] Most of them consist of a xenon atom in the +2, +4, +6, or +8 oxidation state coupled to an extremely electronegative element, such fluorine or oxygen, as in xenon difluoride (XeF).

2), xenon tetrafluoride (XeF

4), xenon hexafluoride (XeF

6), xenon tetroxide (XeO

4), and sodium perxenate (Na

4XeO

6). Xenon reacts with fluorine to form numerous xenon fluorides according to the following equations:

Xe + F2 → XeF2

Xe + 2F2 → XeF4

Xe + 3F2 → XeF6

Some of these compounds have found use in chemical synthesis as oxidizing agents; XeF

2, in particular, is commercially available and can be used as a fluorinating agent.

The molar ratio of (6 moles of NaBr)/(1 mole of Fe₂S₃) is correct based on the balanced chemical equation.

The balanced chemical reaction is given as;

2FeBr₃ + 3Na₂S   →  Fe₂S₃   +  6NaBr

From the balanced chemical reaction above, 6 moles of NaBr will also correspond to 1 mole of  Fe₂S₃.

Thus, the molar ratio of (6 moles of NaBr)/(1 mole of Fe₂S₃) is correct based on the balanced chemical equation.

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Answer:

+3

Explanation:

The phosphate ion has a charge of -3, so for the compound to have an overall change of 0, Fe needs to have a charge of +3.

The  portion of the electromagnetic spectrum do we need to observe is the red part of the This is further explained below.

Generally, Spectrum refers to the range of wavelengths or frequencies that electromagnetic radiation may cover.

In conclusion, We need to keep an eye on the red part of the electromagnetic spectrum since a large positive redshift causes an increase in wavelength and a corresponding fall in frequency and photon energy (such as light).

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1) There are only single bonds between the carbon atoms, so the ending is 'ane'. Thus, the answer is 2-methylbutane.

2) There is one triple bond between one pair of carbon atoms, so the ending is 'yne'. So, the answer is 2-butyne.

3) There is one double bond between one pair of carbon atoms, so the ending is 'ene'. So, the answer is 2-butene.

[tex]density = \frac{mass}{volume} [/tex]

[tex]d = \frac{25.7}{22.6} = 1.13717 \: g/ml[/tex]

[tex]pay \: attention \: to \: the \: unit \: used \: to \\ express \: the \: density \\ since \: the \: mass \: is \: given \: in \: grams \\ and \: volume \: in \: milliliter \: then \\ the \: density \: is \: expressed \: in \\ grams \: per \: \: milliliter[/tex]

because if you take 42-10 = 32

32 is the answer to this question 14x3-2x5.

According to the BODMAS rule, if an expression contains brackets ((), {}, []) we have first to solve or simplify the bracket followed by 'order' (that means powers and roots, etc.), then division, multiplication, addition and subtraction from left to right.

14x3-2x5

42 -10

32

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Answer:

A

Explanation:

Answer:

its D) all of the above

Explanation:

i learned this is like 6th grade

The weaker bond is the bond between the two hydrogen atoms since it has a lower bond energy.

The term bond energy has to do with the energy that is required to break a bond. It is also the energy that must be supplied when a bond is formed.

The lower the bond energy of  bond, the weaker the bond. Hence, it follows that the weaker bond here is the bond between the two hydrogen atoms since it has a lower bond energy.

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When 400g each of water, oil, and sand are taken and heated from room temperature to 700⁰C on identical Bunsen burners. In this case, Substances that take more time to get heated will take more time to cool Hence, Option (A) is correct.

Heat transfer is a form of energy transfer and can occur by conduction, convection, and/or radiation.

Heat transfer occurs any time there is a temperature difference between two objects and occurs in the direction of decreasing temperature, meaning from a hot object to a cold object.

Tests show that the transferred heat relies upon three factors—the adjustment of temperature, the mass of the framework, and the substance and period of the substance.

One of the significant impacts of heat transfer is temperature change: warming builds the temperature while cooling diminishes it.

We expect no stage change and no work is done on or by the framework.

In this case, Substances that take more time to get heated will take more time to cool Hence, Option (A) is correct.

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It will take less 1.8 seconds for the mass of a sample of magnesium-20 to decay from 65.6 micrograms to 8.20 micrograms.

The half-life of a radioactive element is the time for half the amount of a sample of the substance to decay.

After 0.6 seconds 31.25 remains

After 1.2 seconds, 15.625 remains

After 1.8 seconds, 7.9 micrograms remains.

In conclusion, it will take less 1.8 seconds for the sample to decay to 8.20 micrograms.

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The time taken for the radioactive sample to decay is 1.8 s.

The half life is the time taken for only half of the number of the radioactive isotopes to remain. Now we have;

No = 65.6 micrograms

N = 8.20 micrograms

t1/2 = 0.600 seconds

t = ?

Hence;

N/No = (1/2)^t/t1/2

8.20/ 65.6  = (1/2)^t/0.600

0.125 =  (1/2)^t/0.600

1/8 =  (1/2)^t/0.600

(1/2)^3 =  (1/2)^t/0.600

3 = t/0.600

t = 3 * 0.600

t = 1.8 s

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Answer:

8.81 atm

Explanation:

If we let the unknown pressure be x,

[tex] \frac{713 \times 2.81}{664} = \frac{126x}{368} \\ \\ x = 8.81[/tex]

1) The theoretical yield of SnS2 will be 4.20 grams

2) The percent yield will be 7.93%

From the equation of the reaction:

[tex]SnBr_4(aq)+2Na_2S(aq)-- > 4NaBr(aq)+SnS_2(s)[/tex]

The mole ratio of the reactant is 1:2.

Mole of 48.1 mL, 0.478 M SnBr4  = 0.478 x 48.1/100 = 0.023 mols

Mole of 48.8 mL, 0.160 M Na2S = 0.160 x 48.8/1000 = 0.0078 moles

Thus, Na2S is in excess while SnBr4  is limiting.

Mole ratio of SnBr4  and SnS2 = 1:1

Equivalent mole of SnS2 = 0.023 moles

Mass of 0.023 noles SnS2 = 0.023 x 182.81 = 4.20 grams

2) With 0.0333 g of SnS2 recovered, percent yield = 0.333/4.2 x 100 = 7.93%

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Here is the complete question:

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)

Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.

1) Calculate the theoretical yield of SnS2. ;

2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.

Answer:

Neutron does not contain any charge because the charge of the quarks that made up the neutron balances each other out.

Hope it helps.

The pH of the solution is obtained as 2.23.

The pKa shows the extent to which an acid is dissociated in solution. Now we have;

        HCOOH(aq)   +    H2O(l) ⇔  HCOO-(aq)    +  H30+(aq)

Ka = [HCOO-] [ H30+]/[ HCOOH]

But =  [HCOO-] = [ H30+] = x

Ka = x^2/[HCOOH]

x = √Ka [HCOOH]

x = √0.19 *  1.8×10−4

x = 5.8 * 10^-3

pH = - log (5.8 * 10^-3)

pH = 2.23

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The net ionic equation of the reaction is:

Net ionic equations are equations where ions which remain in solution known as spectator ions are not shown in the equation. Only ions involved in formation of product are shown.

In the given equation, sodium and nitrate ions are spectator ions.

The net ionic equation of the given reaction is as follows:

In conclusion, spectator ions are not shown in net ionic equation.

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Answer:

Therefore the theoretical density of iron is 7.877 g/cm³ .

Therefore the number of vacancy per cm³ is

Explanation:

The moles of Na⁺ in 1.30 L is 0.2158 moles of Na⁺.

Concentration of a solution is the amount of solute present in a given volume of solution or solvent.

The concentration of the Na⁺ in the  intravenous saline solution contains 166 mEq/L.

1 mEq/Liter = 1 mmol/L

Moles of Na⁺ in 1.30 L = M

[tex]M = {166 mEq/L} * \frac{1 mmol/L}{1 mEq/L} * \frac{1.30 L}{1000 mL} = 0.2158\:moles[/tex]

Moles of Na⁺ in 1.30 L = 0.2158 moles of Na⁺.

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Answer:

Entropy is a measurement of how much the atoms in a substance are free to spread out, move around, and arrange themselves in random ways

Answer: 6 valence electrons

Explanation: the atomic number for oxygen is 8. the first shell takes 2, the second 8. so, the outer shell is the last shell, which takes 6 from oxygen because oxygen has only 8.  6 is the number for outer or valence electron for oxygen

637,980 Joules is needed to raise the temperature of 2.50 kg of this oil from 23 ∘C to 191 ∘C.

The heat in calories required to raise the temperature of one gram of a substance one degree Celsius.

Q = cmΔT

where

Q = heat energy needed for that material to get desired temperature change (in Joules)

M = mass (in grams) so you have to convert from kilograms.

c = specific heat constant for the material being heated

ΔT = change in temperature

Q = (1.75)(2.17 x 1000)(191 - 23)

Q = (1.75)(2170)(168)

Q = 637,980 Joules

Hence, 637,980 Joules is needed to raise the temperature of 2.50 kg of this oil from 23 ∘C to 191 ∘C.

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The molecular formula of the compound is C6H6

First, convert the percentages to moles

C = 92.9/12 = 7.74

H = 7.71/1 = 7.71

Thus, the empirical formula will be  CH

[empirical formula]n = molecular formula

Empirical formula mass = 12 + 1 = 13

n = empirical formula mass/molar mass = 78/13 6

Thus, the molecular formula will be C6H6

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The beginning and ending energy levels of the electron during the emission of energy that leads to spectral line n(i)= 2 and n(f)=2

wavelength(λ) = 656.3 nm = 656.3 x 10^-9

E= hc / λ

E = energy of the photon

h=Planck's constant

c=speed of light

λ= wavelength of the photon

E= (6.626 x 10^-34)(3 x 10^8)/( 656.3 x 10^-9)

E = 3.028 x 10^-37J

It's a red line which means it falls in a visible region which is known as the Balmer series so;

n (f) = 2

ΔE= RH (1/n(f)^2 - 1/n(i)^2)

3.028 x 10^-37 =-2.18 x 10^-18(1/2^2 - 1/n(i))

n(i)= 2

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0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

Define the molarity of a solution.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Given data:

Volume of solution = 911 mL=0.911L

Given the mass of  MgS = 29.9 g

Molar mass of MgS = 56.38 g/mol

Moles of MgS = 0.5303299042

[tex]Molality = \frac{Moles \;solute}{Volume \;of \;solution \;in \;litre}[/tex]

[tex]Molality = \frac{0.5303299042}{0.911L}[/tex]

Molality = 0.5821403998=0.58

Hence,  0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

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