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The crystal structure of salt is

Answer:

equal to zero is the right answer

Answer:

One structure shows all the bonds whereas the other shows some of the bonds.

Explanation:

Expanded structure shows all of the bonds connecting all of the atoms in the compound while on the hand, condensed structural shows all atoms, but exclude some or all of the vertical and horizontal bonds. condensed structure makes it easier to write the formula in a line as compared to expanded structure. A bond-line representation is a kind of representation of molecular structure of compounds on a line. In this representation, covalent bonds are represented with one line for each level of bond order.

The correct answer is mixture

Answer:

potentiol or kenetic

Explanation:

Answer:

A buffer solution is prepared by adding 13.74 g of sodium acetate (NaC2H3O2) and 15.36 g of acetic acid to enough water to make 500 mL of solution.

Calculate the pH of this buffer.

Explanation:

The pH of a buffer solution can be calculated by using the Henderson-Hesselbalch equation:

[tex]pH=pKa+log\frac{[salt]}{[acid]}[/tex]

The pH of the given buffer solution can be calculated as shown below:

The fluorine atom is partially negatively charged while chlorine is partially positively charged.

Polar molecules are molecules whose molecules are partially charged due to the electronegative differences between the atoms in the molecule of the compound.

Chlorine monofluoride is a polar molecule.

Fluorine is more electronegative than the chlorine atom.

Therefore, the fluorine atom is partially negatively charged while chlorine is partially positively charged.

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Answer:

[tex]m_{CO_2}=75.6gCO_2[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out mandatory for us to calculate the reacting moles of both C and O2 because we are given grams and pressure, temperature and volume, respectively:

[tex]n_C=36gC*\frac{1molC}{12gC}=3.0molC \\\\n_{O_2}=\frac{3.0atm*14L}{0.08206\frac{atm*L}{mol*K}*298K}=1.72molO_2[/tex]

Thus, since C and O2 react in a 1:1 mole ratio, we infer C is in excess, and the grams of CO2 can be calculated with the moles of O2:

[tex]m_{CO_2}=1.72molO_2*\frac{1molCO_2}{1molO_2}*\frac{44.01gCO_2}{1molCO_2} \\\\ m_{CO_2}=75.6gCO_2[/tex]

Best regards!

Answer:

it is...

Explanation: Pure copper(II) sulfate is white. It is also known as anhydrous copper(II) sulfate because it has no water in it. When water is present in a sample of copper(II) sulfate it turns blue.

Answer:

put it in a beaker then tne beaker in a boiling water bath and when it temperature has reached 100°c condense that gas ... if there is no gas evaporating by 100° then there is no water

Explanation:

water evaporates at 100°c

Answer:

a

Explanation:

Answer:

Substance 1 will diffuse at a faster rate.

Explanation:

We can solve this problem by keeping in mind Fick's law, which states:

Where:

As (dc/dx) is equal for both substances, as stated by the problem, the substance with the higher diffusion constant will diffuse at a faster rate.

Thus the answer is substance 1.

Answer:

2.37 (w/w)% of NH3 in the fertilizer

Explanation:

The HCl reacts with NH3 as follows:

HCl + NH3 ⇄ NH4Cl

To solve this question we must find the  moles of HCl used in the titration = Moles NH3. With its molar mass we can find mass of NH3 and using the dilutions we can find weight percent as follows:

Moles HCl = Moles NH3

32.37mL = 0.03237L * (0.1077mol/L) =

Mass NH3 in the dilution -Molar mass: 17.031g/mol-

0.003486moles NH3 * (17.031g/mol) = 0.05937g NH3

Mass NH3 in the sample:

0.05937g NH3 * (79.733g + 24.803g) / 10.560g =

0.588g NH3

Weight percent:

0.588g NH3 / 24.803g * 100 =

Answer:

[tex]v_{solute}=160mL[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the volume of methyl alcohol solute by using the definition of by-volume percentage:

[tex]\%v=\frac{v_{solute}}{v_{solution}} *100\%[/tex]

Whereas we solve for the volume of the solute as shown below:

[tex]v_{solute}=\frac{\%v*v_{solution}}{100\%} \\\\v_{solute}=\frac{80\%*200mL}{100\%}\\\\ v_{solute}=160mL[/tex]

Regards!

Answer:

67 L

Explanation:

Step 1: Write the balanced equation

Na₂S(aq) + 2 HCl(aq) → 2 NaCl(aq) + H₂S(g)

Step 2: Calculate the moles corresponding to 234 g of Na₂S

The molar mass of Na₂S is 78 g/mol.

234 g × 1 mol/78 g = 3.0 mol

Step 3: Calculate the moles of H₂S produced from 3.0 moles of Na₂S

The molar ratio of Na₂S to H₂S is 1:1. The moles of H₂S formed are 1/1 × 3.0 mol = 3.0 mol.

Step 4: Calculate the volume occupied by 3.0 moles of H₂S at STP

At STP, 1 mole of H₂S occupies 22.4 L.

3.0 L × 22.4 L/1 mol = 67 L

Answer:

Al2o3 is 101

(nh4)2O IS 52

S8 is 256.56

Ba(oh)2 is 171.35

Cacl2 is 110.98

H2O is 18.01

Explanation:

Answer:

a) 0.15 mol.

b) 8.95 g.

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to infer this problem is solved by using the ideal gas equation:

[tex]PV=nRT[/tex]

And proceed as follows:

a) Here, we solve for the moles, n,  as follows:

[tex]n=\frac{PV}{RT} \\\\n=\frac{0.50atm*4.5L}{0.08206\frac{atm*L}{mol*K}*178K} \\\\n=0.15mol[/tex]

b) for the calculation of the mass, we recall the molar mass of butane, 58.12 g/mol, to obtain:

[tex]0.15mol*\frac{58.12g}{1mol} =8.95g[/tex]

Regards!

Answer:

1.104 J/g°C

Explanation:

Using Q = m × c × ∆T

Where;

m = mass of substance (g)

c = specific hear capacity (J/g°C)

∆T = change in temperature (°C)

For a colorimeter,

Q(water) = - Q(metal)

m. c. ∆T (water) = - m. c. ∆T (metal)

According to the information provided;

For water:

m = 28.0g

c = 4.184 J/g°C

∆T = (21.23 - 19.73°C)

For the metal:

m = 2.05g

c = ?

∆T = (21.23 - 98.88°C)

m. c. ∆T (water) = - m. c. ∆T (metal)

[28 × 4.184 × (21.23 - 19.73°C)] = -[2.05 × c × (21.23 - 98.88°C)]

[117.152 × 1.5] = -[2.05 × c × (-77.65)]

175.728 = -[-159.1825c]

175.728 = 159.1825c

c = 175.728 ÷ 159.1825

c = 1.104

c = 1.104 J/g°C

Answer:

combine harvester, or a combiner.

Answer:

1.06 °C

Explanation:

From the question given above, the following data were obtained:

Mass of gold (M₉) = 10 g

Specific heat capacity of gold (C₉) = 0.129 J/gºC

Initial temperature of gold (T₉) = 24 °C

Mass of water (Mᵥᵥ) = 118 g

Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Initial temperature of water (Tᵥᵥ) = 1 °C

Equilibrium temperature (Tₑ) =?

The equilibrium temperature of the system can be obtained as follow:

Heat loss by the gold = heat gained by the water

M₉C₉(T₉ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)

10 × 0.129 (24 – Tₑ) = 118 × 4.184 (Tₑ – 1)

1.29(24 – Tₑ) = 493.712 (Tₑ – 1)

Clear bracket

30.96 – 1.29Tₑ = 493.712Tₑ – 493.712

Collect like terms

30.96 + 493.712 = 493.712Tₑ + 1.29Tₑ

524.672 = 495.002Tₑ

Divide both side by 495.002

Tₑ = 524.672 / 495.002

Tₑ = 1.06 °C

Therefore, the temperature of the system is 1.06 °C

The amount of heat of the system is measured by a device called a calorimeter. The final temperature of the system will be 1.06 degrees celsius.

The equilibrium temperature is the temperature that follows the law of thermodynamics and is said to be the system that has alike temperatures.  

Given,

Mass of Ag [tex]\rm (M_{g})[/tex] = 10g

Specific heat capacity of Ag [tex](\rm C_{g})[/tex] = [tex]\rm 0.129 J/g^{\circ}C[/tex]

The initial temperature of Ag [tex](\rm T_{g})[/tex] = [tex]24 ^{\circ}\;\rm C[/tex]

Mass of water [tex](\rm M_{w})[/tex] = 118 g

Specific heat capacity of water [tex](\rm C_{w})[/tex] = [tex]4.184 \rm \;J/g^{\circ}\;\rm C[/tex]

The initial temperature of water [tex](\rm T_{w})[/tex] = [tex]1 ^{\circ}\;\rm C[/tex]

Equilibrium temperature = [tex](\rm T_{e})[/tex]

The equilibrium temperature can be shown as, heat loss by the gold = heat gained by the water:

[tex]\rm \rm M_{g}C_{g}(T_{g} - T_{e}) = M_{w}C_{w}(T_{e}-C_{w})[/tex]

Substituting values in the equation:

[tex]\begin{aligned} 10 \times 0.129 (24 - \rm T_{e}) &= 118 \times 4.184 (\rm T_{e} - 1)\\\\\rm 1.29(24 - T_{e}) &= 493.712 (\rm T_{e} - 1)\\\\524.672 &= 495.002 \;\rm T_{e}\end{aligned}[/tex]

Now divide both the sides by 495.002:

[tex]\begin{aligned} \rm T_{e} &= \dfrac{524.672 }{495.002}\\\\\rm T_{e} &= 1.06 \;^{\circ}\rm C\end{aligned}[/tex]

Therefore, the final temperature of the system is 1.06 degrees celsius.

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Answer:

d. 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Explanation:

Aluminum metal reacts with oxygen gas in a combination reaction that forms a product that coats the metal preventing it from further oxidation: aluminum oxide. Aluminum is a cation with charge 3+ (Al³⁻) and oxide is an anion with charge 2- (O²⁻). Thus, the neutral compound aluminum oxide has the chemical formula Al₂O₃. The unbalanced chemical equation is:

Al(s) + O₂(g) → Al₂O₃(s)

We can balance using the trial and error method. First, we will balance O atoms by multiplying Al₂O₃ by 2 and O₂ by 3.

Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Finally, we get the balanced equation by multiplying Al by 4.

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Organic compround is octane C8H18

Answer:

1 Mole = 84.007 g/mol

Explanation:

Sodium bicarbonate (IUPAC name: sodium hydrogen carbonate), commonly known as baking soda or bicarbonate of soda, is a chemical compound with the formula NaHCO3

Answer:

32.0 kJ

General Formulas and Concepts:

Thermochemistry

Specific Heat Formula: q = mcΔT

Explanation:

Step 1: Define

Identify variables

[Given] m = 1.00 g

[Given] ΔT = 1.48 °C

[Given] c = 21.6 kJ/g °C

[Solve] q

Step 2: Find Heat

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs.

31.968 kJ ≈ 32.0 kJ

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

[tex]rate = k[P]^{2} [Q][/tex]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

[tex]rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }[/tex]

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

[tex]k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4[/tex]

Second row:

2. Rate value:

[tex]rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1[/tex]

3.Third row:

[tex][Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}[/tex]

4. Fourth row:

[tex][P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}[/tex]

Answer:

i didnt understand

Explanation:

Answer:

x = 2 (C₂H₆)

Explanation:

The general formula for alkanes is CₓH₂ₓ₊₂

2x + 2 = 6

Simply solve for n:

2x = 4

x = 2

Answer:

The reactive nucleophile is Ketone.

Explanation:

In organic chemistry, The process of acid - catalyzed aldol condensation starts from when ketone (or any aldehyde) is converted to an -enol, after which it attacks another ketone/aldehyde that has already been activated by parbonyl oxygen protonation.

The process of this is that first of all the ketone undergoes tautomerization to form -enol. Thereafter, the other carbonyl will undergo protonation which makes the carbon activated towards attack. Now, the nucleophilic enol will be added to the carbonyl in a [1,2]-addition reaction and we will now use deprotonation to obtain the neutral Aldol product.

Now, since only the ketone can produce an -enol, thus it is the nucleophile as aldehydes are better electrophiles

Answer:

The student is now told that the four solids, in no particular order, are barium chloride (BaCl2), sugar (C6H12O6), butanoic acid (C3H7COOH), and sodium bromide (NaBr). Assuming that conductivity is correlated to the number of ions in solution, rank the four substances based on how well a 0.20 M solution in water will conduct electricity. Rank from most conductive to least conductive.

Explanation:

The given substances are:

barium chloride(BaCl2),

glucose(C6H12O6),

butanoic acid (C3H7COOH) which is a weak acid,

sodium bromide (NaBr).

The conductivity of a solution is proportional to the number of ions present in a particular solution.

1mol. of BaCl2 in water produces a total three mol. of ions.

[tex]BaCl_2 (aq) -> Ba^2^+(aq) + 2Cl^-(aq)[/tex]

Gluocse is a covalent compound and it does not dissociate into ions in water.

So, it does not conduct electricity.

Butanoic acid is a weak acid. But due to the release of H+ ions it can conduct a very less amount of electricity.

NaBr is an ionic compound and in 1mol. of NaBr in water gives two mol. of ions.

NaBr (aq)  -> Na+ (aq)  + Br- (aq)

Hence, the order of conductivity among the given substances in aqueous solution is:

BaCl2 > NaBr > butanoic acid > glucose

Answer:

Explanation:

578kj/mol