The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. How much tritium will be left after 30 years

Answer: The lowest possible melting point for engine coolant that is 26.1% ethylene glycol is [tex]-7.83^0C[/tex]

Explanation:

Depression in freezing point or melting point is given by:

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

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

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

[tex]K_f[/tex]  = freezing point constant for water= [tex]1.86^0C/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]

Given : 26.1 g of ethylene glycol in 100 g of water

Weight of solvent (water)= 100 g = 0.1 kg     (1kg=1000g)

Molar mass of solute (ethylene glycol)  = 62 g/mol

Mass of  solute (ethylene glycol) added = 26.1 g

[tex](0-T_f)^0C=1\times 1.86\times \frac{26.1g}{62g/mol\times 0.1kg}[/tex]

[tex](0-T_f)^0C=7.83[/tex]

[tex]T_f=-7.83^0C[/tex]

Thus  the lowest possible melting point for engine coolant that is 26.1% ethylene glycol is [tex]-7.83^0C[/tex]

Answer:

[tex]C_2=0.97M[/tex]

Explanation:

Hello,

In this case, for dilution process, we can notice that the initial moles remain the same once the dilution is completed, therefore, both concentration and volume change considering:

[tex]n_1=n_2\\\\V_1C_1=V_2C_2[/tex]

In such a way for the given final volume, the resulting concentration is noticed to be:

[tex]C_2=\frac{V_1C_1}{V_2} =\frac{10mL*2.5M}{25.8mL}\\ \\C_2=0.97M[/tex]

This is supported by the fact that the higher the volume the lower the concentration.

Best regards.

Answer:

C7H5F3

Explanation:

The following data were obtained from the question:

Mass of Carbon (C) = 57.53g

Mass of Hydrogen (H) = 3.45g

Mass of Fluorine (F) = 39.01g

The empirical formula of the compound can be obtained as follow:

C = 57.53g

H = 3.45g

F= 39.01g

Divide each by their molar mass

C = 57.53/12 = 4.79

H = 3.45/1 = 3.45

F = 39.01/19 = 2.05

Divide each by the smallest

C = 4.79/2.05 = 2.3

H = 3.45/2.05 = 1.7

F = 2.05/2.05 = 1

Multiply through by 3 to express in whole number

C = 2.3 x 3 = 7

H = 1.7 x 3 = 5

F = 1 x 3 = 3

Therefore, the empirical formula for the compound is C7H5F3

Answer:

1.41 M

Explanation:

First we must use the information provided to determine the concentration of the aluminum hydroxide.

Mass of aluminum hydroxide= 320mg = 0.32 g

Molar mass of aluminum hydroxide= 78 g/mol

Volume of the solution= 5.00 ml

From;

m/M= CV

Where;

m= mass of aluminum hydroxide= 0.32 g

M= molar mass of aluminum hydroxide = 78 g/mol

C= concentration of aluminum hydroxide solution = the unknown

V= volume of aluminum hydroxide solution = 5.0 ml

0.32 g/78 g/mol = C × 5/1000

C = 4.1×10^-3/5×10^-3

C= 0.82 M

Reaction equation;

Al(OH)3(aq) + 3HCl(aq) -----> AlCl3(aq) + 3H2O(l)

Concentration of base CB= 0.82 M

Volume of base VB= 1.60 ml

Concentration of acid CA= the unknown

Volume of acid VA= 2.80 ml

Number of moles of acid NA = 3

Number of moles of base NB= 1

Using;

CA VA/CB VB = NA/NB

CAVANB = CBVBNA

CA= CB VB NA/VA NB

CA= 0.82 × 1.60 × 3/ 2.80 ×1

CA= 1.41 M

Therefore the concentration of HCl is 1.41 M

Answer:

The gas evolved because of reaction of acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas. This happens because of formation of white precipitate of calcium carbonate.

Hope it helps you!

The milky substance formed is CO₂ gas.

The gas evolved because of reaction of acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas.

This happens because of formation of white precipitate of calcium carbonate.

Lime water turns milky when the gas liberated from an acidified carbonate solution is passed into it.

The liberated CO₂ reacts with lime water to give calcium bicarbonate as the precipitate.

Hence, the milky substance formed is CO₂ gas.

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Answer:

HPO₄⁻² predominates at pH 9.3

Explanation:

These are the equilibriums of the phosphoric acid, a tryprotic acid where 3 protons (H⁺) are realesed.

H₃PO₄ + H₂O   ⇄   H₂PO₄⁻   +   H₃O⁺   pKa 2.14

H₂PO₄⁻  +  H₂O   ⇄   HPO₄⁻²  +  H₃O⁺  pKa 6.86

HPO₄⁻²  +  H₂O   ⇄   PO₄⁻³   +   H₃O⁺  pKa 12.38

The H₂PO₄⁻  works as amphoterous, it can be a base and acid, according to these equilibriums.

H₂PO₄⁻ +  H₂O   ⇄   HPO₄⁻²  +  H₃O⁺

H₂PO₄⁻ +  H₂O   ⇄   H₃PO₄  +  OH⁻

pH 9.3 is located between 6.86 and 12.38 where we have this buffer system HPO₄⁻² / PO₄⁻³, where the HPO₄⁻² is another amphoterous:

HPO₄⁻ +  H₂O   ⇄   H₂PO₄⁻  +  OH⁻

HPO₄⁻²  +  H₂O   ⇄   PO₄⁻³   +   H₃O⁺

The media from the two pKa, indicates the pH where the protonated form is in the same quantity as the unpronated form, so:

(6.86 + 12.38) /2 =  9.62

Above this pH, [PO₄⁻³] > [HPO₄⁻²].

In conclussion, at pH 9.3, [HPO₄⁻²] > [PO₄⁻³]

Answer:

Joseph Henry.

Explanation:

Hello,

In this case, we define the accuracy based on the degree of deviation a measure has regarding the the true or expected value. In this case, the lower the difference, the more accurate the procedure. In such a way, for the two students, we compute the difference between the expected value and the obtained value:

[tex]Mary=2.300g/mL-2.226g/mL=0.074g/mL[/tex][tex]Joseph=2.335g/mL-2.300mL=0.035g/mL[/tex]

In such a way, since the difference is lower in Joseph's measure, we can say he is the most accurate.

Best regards.

Answer:

1661μL of a 6M HCl you need to add

Explanation:

pH is defined as -log[H⁺] ([H⁺] =10^{-pH}), the initial and final concentrations of [H⁺] you need are:

Initial [H⁺] = 10^{-3.5} = 3.16x10⁻⁴M H⁺

Final [H⁺] = 10^{-1} = 0.1M H⁺

In moles, knowing volume of the solution is 0.1L:

Initial [H⁺] = 0.1L ₓ (3.16x10⁻⁴mol H⁺ / L) = 3.16x10⁻⁵moles H⁺

Final [H⁺] = 0.1L ₓ (0.1mol H⁺ / L) = 0.01 moles H⁺.

That means, moles of H⁺ you need to add to the solution is:

0.01mol - 3.16x10⁻⁵moles = 9.9684x10⁻³ moles of H⁺.

A solution of HCl dissociates in H⁺ and Cl⁻ ions, that means moles of HCl added are equal to moles of H⁺. As you need to add 9.9684x10⁻³ moles of H⁺ = 9.9684x10⁻³ moles of HCl:

9.9684x10⁻³ moles of HCl ₓ (1L / 6mol) = 1.6614x10⁻³L

In μL:

1.661x10⁻³L × (1x10⁶μL / 1L) =

Answer:

The question is incomplete, the solute was not given.

Let the solute be K₂CrO₄ and the solvent be water

Complete Question should be like this:

The density of a 0.438 M solution of potassium chromate (K₂CrO₄) at 298 K is 1.063 g/mL.

Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation.  

Pvap = ________atm

Answer:

Pvap (of water above the solution) = 0.0306 atm

Dissolution of the solute

K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻

Explanation:

Given

volume of solution = 1 Litre = 1000 mL of the solution

density of the solution = 1.063 g/mL

concentration of the solution= 0.438M

temperature of the solution= 298 K

vapour pressure of pure water = 0.0313atm

Recall: density = mass/volume

∴mass of solution = volume x density

m = 1000 x 1.063 = 1063 g

To calculate the moles of K₂CrO₄ = volume x concentration

= 1 x 0.438 = 0.438 mol

Mass of K₂CrO₄ = moles x molar mass = 0.438 x 194.19 = 85.055 g

Mass of water = mass  of solution - mass of K₂CrO₄

= 1063 - 85.055 = 977.945 g

moles of water = mass/molar mass

∴ moles of water = 977.945/18.02 = 54.27 mol

 Dissolution of the solute

K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻

(dissolution is the process by which solute(K₂CrO₄) is passed into solvent(H₂O) to form a solution

moles of ions = 3 x moles of K₂CrO₄

= 3 x 0.438 = 1.314 mol

Vapor pressure of solution = mole fraction of water x vapor pressure of water  

= 54.27/(54.27 + 1.314) x 0.0313 = 0.0306 atm

Answer:

[tex]m_{Fe}=23.0gFe[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]FeO+Mg\rightarrow Fe+MgO[/tex]

Thus, for the given masses of reactants we should compute the limiting reactant for which we first compute the available moles of iron (II) oxide:

[tex]n_{FeO}=40.0gFeO*\frac{1molFeO}{72gFeO} =0.556molFeO[/tex]

Next, we compute the consumed moles of iron (II) oxide by the 10.0 g of magnesium, considering their 1:1 molar ratio in the chemical reaction:

[tex]n_{FeO}^{consumed}=10.0Mg*\frac{1molMg}{24.3gMg}*\frac{1molFeO}{1molMg}=0.412molFeO[/tex]

Therefore, we can notice there is less consumed iron (II) oxide than available for which it is in excess whereas magnesium is the limiting reactant. In such a way, the produced mass of iron turns out:

[tex]m_{Fe}=0.412molFeO*\frac{1molFe}{1molFeO}*\frac{56gFe}{1molFe}\\ \\m_{Fe}=23.0gFe[/tex]

Regards.

Answer:

Mg, Sc, Ca

Explanation:

To figure out increasing atomic radii, we use Periodic Trends applied to the Elements of the Periodic Table to help us out. We know that the trend for atomic radii is increasing left and down. Since Ca is the furthest down and left of the 3, it has the largest atomic radius. Since Sc is next element to Ca, it would be the 2nd largest atomic radius of the 3. Since Mg is above Ca, it has the smallest atomic radius of the 3.

Answer:

B.) 2x

Explanation:

Hello,

In this case, we can apply the following rule of three, knowing that 0.1 dm³ equals x and 0.2 dm³ is the unknown:

[tex]0.1dm^3\longrightarrow x\\0.2dm^3 \longrightarrow ?[/tex]

Thus, solving for the unknown we find:

[tex]?=\frac{0.2dm^3*x}{0.1dm^3} \\\\?=2*x[/tex]

Therefore, the answer is B.) 2x.

Best regards.

Answer:

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,  

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,  

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.  

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,  

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,  

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,  

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,  

x as 0.115 and y as 0.165

Now the mass of AgCl will be,  

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,  

278.1 × 0.165 = 45.88 grams.  

Answer:

Check Explanation.

Explanation:

Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.

NaBr (s)

The Standard formation reaction is

Na (s) + (1/2)Br₂ (g) → NaBr (s)

Using appendix C, the standard heat of formation of NaBr(s) is

ΔH∘f = -359.8 kJ/mol.

SO₃ (g)

The Standard formation reaction is

S (s) + (3/2) O₂ (g) → SO₃ (g)

Using appendix C, the standard heat of formation of SO₃(g) is

ΔH∘f = -395.2 kJ/mol.

Pb(NO₃)₂ (s)

The Standard formation reaction is

Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)

Using appendix C, the standard heat of formation of Pb(NO₃)₂(s) is

ΔH∘f = -451.9 kJ/mol.

Hope this Helps!!!

Answer:

Explanation:

Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.

Answer:

The ratio of the mass of an adult rhino to the mass of a small egret is

(1400 kg)/(0.76 kg) = 1842.11

The ratio of the mass of a proton to the mass of an electron is

1836.15

The two ratios are very similar, the ratio of the masses of the proton to the electron being slightly smaller.

Answer:

Explanation:

0,44

Answer:

14.6 g

Explanation:

Fe₂O₃  +  3 C  →  2 Fe  +  3 CO₃

First, convert grams of Fe₂O₃ to moles.  The molar mass is 159.69 g/mol.

(64.6 g)/(159.69 g/mol) = 0.4045 mol

Now, use stoichiometry to convert moles of Fe₂O₃ to moles of C.  By looking at the chemical equation, you can see that for every 1 mole of Fe₂O₃, you need 3 moles of C.  Use this relationship.

(0.4045 mol Fe₂O₃) × (3 mol C/1 mol Fe₂O₃) = 1.214 mol C

Now, convert moles of C to grams.  The molar mass is 12.01 g/mol.

(1.214 mol C) × (12.01 g/mol) = 14.58 g

You need to use significant figures.

14.58 g ≈ 14.6 g

Answer:

Ecell = +0.25V

Explanation:

the half-cell reactions for a voltanic cell

cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)

anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻

we have the standard cell potential E⁺cell = 0.18V at 80C respectively

Q = [H⁺]/[Cl⁻]

sub for [H+] = 0.10M and [Cl-] = 1.5M

Q= 0.1M/1.5M

Q = 0.067

Ecell = E⁺cell - [tex]\frac{0.059}{n}[/tex] logQ

= 0.18 - [tex]\frac{0.056}{1}[/tex] log 0.067

0.18- 0.059(-1.174)

Ecell = +0.25V

Answer:

Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)

Explanation:

When lead (II) nitrate, Pb(NO3)2 undergoes a double displacement reaction with aqueous sodium chloride, NaCl, the following products are obtained:

Pb(NO3)2(aq) + NaCl(aq) —>

Aqueous Pb(NO3)2 will dissociate in solution as follow:

Pb(NO3)2(aq) —> Pb2+(aq) + 2NO3-(aq)

On the other hand, aqueous NaCl will dissociate as follow:

NaCl(aq) —> Na+(aq) + Cl-(aq)

The double displacement reaction will take place as follow:

Pb(NO3)2(aq) + NaCl(aq) —>

Pb2+(aq) + 2NO3-(aq) + Na+(aq) + Cl-(aq) —> Pb2+(aq) Cl-(aq) + Na+(aq) 2NO3-(aq)

Pb(NO3)2(aq) + NaCl(aq) —> PbCl2(s) + NaNO3(aq)

We simply balance the equation by putting 2 in front of NaCl and 2 in front of NaNO3 as shown below:

Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)

Answer:

BF3

Explanation:

Hybridization can be defined as the mixing of two or more atomic pure orbitals. ( s, p ,  and d) to produce two or more hybrid atomic orbitals that are similar and identical in shape and energy e.g sp,sp²,sp³ ,sp³d, sp³d². Usually , the central atom of a covalent molecules or ion undergoes hybridization.

in BF3; Boron is the central atom. Here, A 2s  electron is excited from the ground state of boron ( 1s²2s²2p¹) to one empty orbitals of 2p.

The 2s orbital is then mixed with two orbitals of 2p to form three sp² hybrid orbitals tat are trigonally arranged in the plane in order to minimize repulsion . Each of the three hybrid orbitals overlaps with p-orbital of fluorine atom to form three bonds of equal strength and with bond angles of 120⁰.

                                  Energy                                                                  

B     ⇵  ║   ⇅  ║    ↑     ----------->    *B    ⇵ ║     ↑  ║     ↑  ║   ↑

       1s      2s         2p                              1s       2s           2p  

    Ground state                                        Excited State

The above shows an illustrative example of how electrons  move from the ground state to the excited state.

Answer:

First, write a balanced equation of the reaction.

A hydrocarbon reacting with oxygen usually gives out carbon dioxide and water (combustion)

2C2H6 + 7O2 ---> 4CO2 + 6H2O

From the equation, we can see the mole ratio of ethane : oxygen is 2:7, meaning 2 moles of ethane reacts with 7 moles of oxygen.

If there's 1 mole of ethane gas, we need y moles of oxygen gas for complete reaction.

[tex]\frac{2}{7} =\frac{1}{y}[/tex]

y= 3.5 moles

Since 5 moles of oxygen is more than the required 3.5 moles, we can deduce oxygen gas is in excess, meaning ethane is limiting.

(You can get the same results too if you take y as ethane gas required).

The no. of moles of product produced ALWAYS depend on the no. of moles of the limiting reactant, since they are the ones which reacts completely.

So, from the equation, since the mole ratio of ethane : water = 2:6,

hence, (take the no. of moles of water produced as z),

[tex]\frac{2}{6} =\frac{1}{z} \\z= 3[/tex]

Therefore, 3 moles of water is produced.

Ethane is the limiting reagent in the given reaction. The number of moles of water produced from the reaction is 3 moles.

A limiting reagent can be defined as that reactant in the chemical reaction which is consumed first among the other reactants during the completion of a chemical reaction.

The limiting reagent is the reactant which decides the yield of the product when the quantity of the reactants is not taken in stoichiometry.

Given, chemical reaction of ethane and oxygen is:

[tex]C_2H_6 + \frac{7}{2} O_2 \longrightarrow 2CO_2 +3H_2O[/tex]

Given, the number of moles of ethane = 1 mol

The number of moles of oxygen = 5 mol

One more of ethane reacts with moles of oxygen = 3.5 mol

Therefore, ethane is the limiting reagent in this reaction as the oxygen is given in excess.

From the balanced equation, one mole of ethane produces moles of water equal to 3 moles.

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Answer:

0.49 mol

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCI ⇒ MgCl₂ + H₂

Step 2: Calculate the moles corresponding to 12 g of Mg

The molar mass of Mg is 24.31 g/mol.

[tex]12g \times \frac{1mol}{24.31g} = 0.49mol[/tex]

Step 3: Calculate the moles of H₂ produced by 0.49 moles of Mg

The molar ratio of Mg to H₂ is 1:1. The moles of H₂ produced are 1/1 × 0.49 mol = 0.49 mol.

Answer:

Manganese (II) ion, Mn²⁺

Explanation:

Hello,

In this case, given the overall reaction:

[tex]2Ce^{4+}(aq) + Tl^+(aq) + Mn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Tl^{3+}(aq) + Mn^{2+}(aq)[/tex]

Thus, since manganese (II) ion, Mn²⁺ is both at the reactant and products, we infer it is catalyst, since catalysts are firstly consumed but finally regenerated once the reaction has gone to completion. Moreover, since inner steps are needed to obtain it, we can infer that the given rate law corresponds to the slowest step that is related with the initial collisions between Ce⁴⁺ and Mn²⁺

Best regards.

Answer:

[tex]t=2.08s[/tex]

Explanation:

Hello,

In this case, for first order reactions, we can use the following integrated rate law:

[tex]ln(\frac{[A]}{[A]_0} )=kt[/tex]

Thus, we compute the time as shown below:

[tex]t=-\frac{ln(\frac{[A]}{[A]_0} )}{k}=- \frac{ln(\frac{0.220M}{0.690M} )}{0.55s^{-1}} \\\\t=-\frac{-1.14}{0.550s^{-1}}\\ \\t=2.08s[/tex]

Best regards.

Hey there!

For this we can use the combined gas law:

[tex]\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}[/tex]

We are only working with pressure and temperature so we can remove volume.

[tex]\frac{P_{1} }{T_{1}} = \frac{P_{2} }{T_{2}}[/tex]

P₁ = 2 atm

T₁ = 27 C

P₂ = 2.2 atm

Plug these values in:

[tex]\frac{2atm}{27C} = \frac{2.2atm}{T_{2}}[/tex]

Solve for T₂.

[tex]2atm = \frac{2.2atm}{T_{2}}*27C[/tex]

[tex]2atm * T_{2}={2.2atm}*27C[/tex]

[tex]T_{2}={2.2atm}\div2atm*27C[/tex]

[tex]T_{2}=1.1*27C[/tex]

[tex]T_{2}=29.7C[/tex]

Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.

Hope this helps!

Answer:

Oxidation by FAD  

Explanation:

1. Oxidation by NAD⁺

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻;                  E°´ =  -0.031 V  

NAD⁺ + 2H⁺ + 2e⁻ ⇌ NADH + H⁺;                      E°´ = -0.320 V

Succinate + NAD⁺ ⇌ Fumarate  + NADH + H⁺; E°' =  -0.351 V

2. Oxidation by FAD

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻;        E°´ = -0.031 V  

FAD + 2H⁺ + 2e⁻ ⇌ FADH₂;                    E°´  = -0.219 V

Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V

Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.

FAD is the stronger oxidizing agent.

The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.  

Answer:

density = 5.1g/m³

Explanation:

use ideal gas equation

Pv=nRT

28.26 x 3.51 = m/v x 0.08206 x 237

m/v = (28.26 x3.51)/(237 x 0.08206) = 5.1g/m³

Answer:

Gaseous

Explanation:

Gasses can move freely and do not form the shape of their containers

Liquids are more free than solids, but they conform to the shape of their container

Solids are not free