Tell me, how the different states of matter are different and the same.

Answer:

No, compound A and B are not the same compound

Explanation:

According to the law of definite proportion "every chemical compound contains fixed and constant proportions (by mass) of its constituent elements." (Encyclopedia Britannica)

We can see in the question that the ratio of flourine to sulphur in compound A is 1.18 while the ratio of flourine to sulphur in compound B is 2.37.

The two chemical compounds do not contain a fixed proportion by mass of their constituent elements therefore, they can not be same compound according to the law of definite proportions.

Answer:

0.03947 atm

Explanation:

The relationship between mmHg and atm is given as;

1 atm = 760 mmHg

x atm = 30 mmHg

Upon solving for x;

x * 760 = 30 * 1

x = 30 / 760 = 0.03947 atm

Answer:

Highest temperature which was recorded in Texas was 40 degrees.

Explanation:

Texas has variating weather. In summer Texas city turns to be very hot and during winters the city of Texas experience fall in temperature. There is very few rainfall in Texas due the climatic conditions.

Answer:

true............,....

Answer:False

Explanation:

This statement is false. Correct statement: Photosynthesis produces glucose and oxygen. ... In this process, plants produce glucose and oxygen from carbon dioxide and water in the presence of sunlight. Photosynthesis converts light energy into chemical energy.

Answer:

The correct answer is - 73.64 minutes.

Explanation:

We know that each gram of protein and carbohydrate produces 4 calories of energy whereas fat produces 9 cal energy per gram.

So, the Calories from carbohydrates

= 181 ×4

= 724 Cal

and, 27 g fat produces = 27 × 9 = 243 Cal

and, 16 g protien = 16 × 4 = 64 Cal

so the total energy will be - 243+ 724+ 64 = 1031 Cal

Thus, the time Fred need to run = 1031/14 =73.64 min

Answer:

78.4

Explanation:

Answer:

The answer to this is solids!

Answer:

The oxidation state of the reducing agent has changed from 0 to +2.

Explanation:

reducing agent is anything that loses electron or gains oxygen

in this case, zinc

Answer:

B. Farming

Explanation:

Answer:

Copper, Water, and Oxygen

Explanation:

Because they are compounds and compounds are pure substances <3

Answer: copper

Explanation: because all of the atoms that make up copper are copper i.e its not a compound or mixture. not made od molecules

Answer:

MM = 225.11 g/mol

Explanation:

In this case, let's analyze the data.

We have a 15.2 mg of an unknown electrolyte in 150 mL of solution. The osmotic pressure is 8.44 Torr, and we want the molecular mass of the unknown.

The osmotic pressure can be calculated using the following expression:

π = CRT   (1)    

π: osmotic pressure (8.44 Torr * 1 atm / 760 Torr = 0.011 atm)

C: Concentration of the unknown in the solution

R: universal constant of gases (0.082 L atm / K mol)

T: temperature in Kelvin (25 + 273 = 298 K)

From this expression, we can either solve for C, and then use another expression to calculate the molecular mass, or we can just replace the expressions in the above formula, to get the direct molecular mass. In this case, we'll follow the second method.

Concentration or molarity of a substance can be calculated using:

C = moles / V  (2)

And moles can be calculated using this expression:

moles = m / MM   (3)

Replacing (3) in (2), and then in (1) we have:

C = m / MM * V

π = m * RT / MM * V   (4)

and now, we can solve for MM:

MM = mRT / π V  (5)

Now, we just need to replace the given data into the above expression to get the value of the molecular mass:

MM = (15.2 / 1000) * 0.082 * 298 / 0.011 * 0.150

Hope this helps

Answer:

Rock In molten form is Lava

^_^

Answer:

Tellurium Fluoride

Explanation:

Solution :

Structure I

The formal charge on both Carbon (C) atom is = 4 valance [tex]$e^-$[/tex] - bonds = 0

Formal charge (O) = 6 V.E - 2 bonds - 4 non bonding electrons = 0

Formal charge on (N) = 5 V.E - 3 bonds - 2 non bonding electrons = 0

F.C. on H = 1 V.E. - 1 bond = 0

Overall charge on the molecule = 0 charge

Structure II

Formal charge on both C atom = 4 valence [tex]$e^-$[/tex] - 4 bonds = 0

Formal charge (O) = 6 V.E. - 1 bonds - 6 non bonding electrons = -1 charge

Formal charge on (N) = 5 V.E. - 4bonds - 0 non bonding electrons = +1 charge

F.C on H = 1 V.E. - 1 bond = 0

Overall charge on the molecule = +1 -1

                                                     = 0 charge

Answer:

11.7

Explanation:

The pH is the negative logarithm of the concentration of H+ ions. If the concentration is 2×10-¹² the the pH will be -log(2×10-¹²) which is 11.698 which can be round up to 11.7.

Answer:

Due to displacement of water.

Explanation:

If the water is displaced in the cylinder it means that the gas is produced in the reaction while on the other hand, if the water is not displaced in the cylinder so it means that the reaction does not produces any gas yet. So we can say that the displacement of waters indicates or shows us whether the gas is produced in the reaction or not.

Answer:

[tex]2Na + Ca^{2+} \rightarrow Ca + 2Na^+[/tex]

Explanation:

Hello!

In this case, since this a reaction in which the oxidation state of the metals is changing, we need to apply a redox method for which:

[tex]Na\rightarrow Na^++e^-\\\\Ca^{2+}\rightarrow Ca+2e^-\\\\\\2Na\rightarrow 2Na^++2e^-\\\\Ca^{2+}\rightarrow Ca+2e^-[/tex]

Thus, the result is:

[tex]2Na + Ca^{2+} \rightarrow Ca + 2Na^+[/tex]

Regards!

Answer:

ıt is the projesi where the polen of one flower reaches the sığmaz of another flower.

Answer:

B

Explanation:

Have a nice day☺gsjsjamsvwksk

Answer:

d. 7.7x10^23 molecules

Explanation:

Given the following data:

Mass of water (H2O) = 23g

To find the number of molecules;

First of all, we would determine the number of moles;

[tex] Number \; of \; moles = \frac {mass}{molar mass} [/tex]

Molar mass of water (H2O) = (1 * 2) + 16 = 18 g/mol

Substituting into the equation, we have

[tex] Number \; of \; moles = \frac {23}{18} [/tex]

Number of moles = 1.2778 moles

Now, to find the number of water molecules;

We know that Avogadro constant is equal to 6.02 * 10^23 mol¯¹

Number of water molecules = number of H2O moles * Avogadro constant

Substituting into the equation, we have;

Number of water molecules = 1.2778 × 6.02 * 10^23

Number of water molecules = 7.7 × 10^23 atoms.

Answer:

See explanation

Explanation:

The term ion product is an expression that does not necessarily reflect equilibrium conditions which can be used to determine the concentration of ions in a solution of the stated substance. As a matter of fact, the ion product is different from the solubility product which reflects equilibrium conditions.

The ion product of the following substances are shown below;

magnesium carbonate - MgCO3

MgCO3(aq) -------> [Mg^2+] [CO3^2-]

calcium phosphate - Ca3(PO4)2

Ca3(PO4)2(aq) --------> [Ca^2+]^3  [PO4^3-]^2

silver sulfide - Ag2S

Ag2S(aq) --------> [Ag^+]^2  [S^2-]

Answer:

A liquid freezes into a solid or any other state change is a physical change. Bubbles are produced upon mixing two solutions is a Chemical change as gas formed. a precipitate is formed from two solutions is a Chemical change  The conversion of a substance to another compound over time leads to a color change of the substance due to the insolubility of the compound.

It is a chemical change or properties. A solution that heats up upon mixing with another is a Chemical change known as an exothermic reaction. dissolving the substance in a liquid is a physical change physical that can be regained by evaporation or crystalization.

Answer:

No, you can not calculate the solubility of X in water at 17 0C.

Explanation:

Solubility refers to the amount of a substance that dissolves in 1000 L of water.

To calculate the solubility of a solute in water, all the water is evaporated and the solid is carefully collected, washed, dried and weighed. The mass of solid obtained can now be used to calculate the solubility of the solute in water as long as there was no loss in mass of solid during the experiment.

In this case, the student threw away part of the solid that precipitated. As a result of this, the mass of solid obtained at the end of the experiment is not exactly the total mass of solute that dissolved in the solvent. Hence, the solubility of X in water at 17 0C can not be accurately calculated.

Answer:

i) ∴∫[tex]\frac{dA}{A}[/tex]  = ∫ [tex]\frac{-1}{10} dt[/tex] ------ ( 1 )

ii) C = 4.303

Explanation:

Given data:

water in tank = 10 liters

concentration of salt flowing in = 0.5 liters/min

10 grams of salt is left in tank after 20 minutes

mixed water is drained at 1 liter per minute

i) Adequate mathematical model for the scenario

lets assume ; A( t ) amount of salt after t minute

A( t ) = 10grams   where t = 20 minutes

differentiate A(t)

dA / dt = ( rate in ) - ( rate out )

           = 0.5 * 0  - A/10

∴∫[tex]\frac{dA}{A}[/tex]  = ∫ [tex]\frac{-1}{10} dt[/tex] ------ ( 1 )

ii) hence;  In A = [tex]\frac{-1}{10} t + C[/tex]  -----  ( 2 )

at  t = 20 , A = 10grams

find C = In 10 + 2

          = 2.303 + 2 = 4.303

back to ( 2 )

In A = [tex]\frac{-t}{10} + 4.303[/tex]

∴ A = e [tex]\frac{-t}{10} + 4.303[/tex]

Answer:

Explanation:

Mol weight of HCl = 36.5

.90 g of HCl = .90 / 36.5 mole = .02465 mole

HCl      =    H⁺        +      Cl⁻

1 mole     1 mole

1 mole of HCl gives 1 mole of H⁺

.02465 mole of HCl will give .02465 mole of H⁺

.02465 mole of H⁺ in 5 litre soln

concentration of H⁺ =  .02465 / 5 = 4.93 x 10⁻³ M

pH = - log ( 4.93 x 10⁻³ )

= 3 - log 4.93

= 3 - .693

= 2.307

= 2.31

NaOH :

Mol weight of NaOH  = 40

.50 g of HCl = .50 / 40 mole = .0125 mole

NaOH     =    Na⁺        +      OH⁻

1 mole     1 mole

1 mole of NaOH gives 1 mole of OH⁻

.0125 mole of NaOH will give .0125 mole of OH⁻

.0125 mole of OH⁻ in 4 litre soln

concentration of OH⁻ =  .0125 / 4 = 3.125 x 10⁻³ M

pOH = - log ( 3.125 x 10⁻³ )

= 3 - log 3.125

= 3 - .495

= 2.505

pH = 14 - pOH

= 14 - 2.505

= 11.495 .

= 11.50

Answer:

Explanation:

TiO2 + C + 2 CI2 → TiCI4 + CO2

(a)

(2.16 mol TiO2) x (2 mol CI2 / 1 mol TiO2) x ( 70.9064 g C/mol) = 306 g CI2

(b)

(2.16 mol TIO2) x (1 mol C / 1 mol TiO2) x ( 12.01078 g C/mol)n= 25.9 g C

(c)

Use the law of conservation of mass:

(2.16 mol TiO2) x ( 79.8660 g TiO2/mol) + (306 g CI2 + 25.9 g C = 504 g total.

Hope this help!:)

Answer:

True

Explanation:

Answer:

Approximately [tex]4.92[/tex].

Explanation:

Initial volume of the solution: [tex]V = 190.0\; \rm mL = 0.1900\; \rm L[/tex].

Initial quantity of [tex]\rm NH_3[/tex]:

[tex]\begin{aligned} n({\rm NH_3}) &= c({\rm NH_3}) \cdot V({\rm NH_3}) \\ &= 0.3733\; \rm mol \cdot L^{-1} \times 0.1900\; \rm L \\ &\approx 0.154375\; \rm mol\end{aligned}[/tex].

Ammonia [tex]\rm NH_3[/tex] reacts with hydrochloric [tex]\rm HCl[/tex] acid at a one-to-one ratio:

[tex]\rm NH_3 + HCl \to NH_4 Cl[/tex].

Hence, approximately [tex]n({\rm HCl}) = 0.154375\; \rm mol[/tex] of [tex]\rm HCl\![/tex] molecules would be required to exactly react with the [tex]\rm NH_3\![/tex] in the original solution and hence reach the equivalence point of this titration.

Calculate the volume of that [tex]0.3733\; \rm mol \cdot L^{-1}[/tex] [tex]\rm HCl[/tex] solution required for reaching the equivalence point of this titration:

[tex]\begin{aligned}V({\rm HCl}) &= \frac{n({\rm HCl})}{c({\rm HCl})} \\ &\approx \frac{0.154375\; \rm mol}{0.3733\; \rm mol \cdot L^{-1}} \approx 0.413541\; \rm L\end{aligned}[/tex].

Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately [tex]0.413541\; \rm L + 0.1900\; \rm L \approx 0.6035\; \rm L[/tex].

If no hydrolysis took place, [tex]0.154375\; \rm mol[/tex] of [tex]\rm NH_4 Cl[/tex] would be produced. Because [tex]\rm NH_4 Cl\![/tex] is a soluble salt, the solution would contain [tex]0.154375\; \rm mol\![/tex] of [tex]\rm {NH_4}^{+}[/tex] ions. The concentration of [tex]\rm {NH_4}^{+}\![/tex] would be approximately:

[tex]\begin{aligned}c({\rm {NH_4}^{+}}) &= \frac{n({\rm {NH_4}^{+}})}{V({\rm {NH_4}^{+}})}\\ &\approx \frac{0.154375\; \rm mol}{0.6035\; \rm L} \approx 0.255782\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

However, because [tex]\rm NH_3 \cdot H_2O[/tex] is a weak base, its conjugate [tex]\rm {NH_4}^{+}[/tex] would be a weak base.

[tex]\begin{aligned}pK_{\rm a}({{\rm NH_4}}^{+}) &= pK_{\rm w} - pK_{\rm b}({\rm NH_3})\\ &\approx 13.99 - 4.75 = 9.25\end{aligned}[/tex].

Hence, the following reversible reaction would be take place in the solution at the equivalence point:

[tex]\rm {NH_4}^{+} \rightleftharpoons NH_3 + H^{+}[/tex].

Let [tex]x\; \rm mol \cdot L^{-1}[/tex] be the increase in the concentration of [tex]\rm H^{+}[/tex] in this solution because of this reversible reaction. (Notice that [tex]x \ge 0[/tex].) Construct the following [tex]\text{RICE}[/tex] table:

[tex]\begin{array}{c|ccccc} \textbf{R}& \rm {\rm NH_4}^{+} & \rightleftharpoons & {\rm NH_3}& + & {\rm H}^{+}\\ \textbf{I} & 0.255782 \; \rm M \\ \textbf{C} & -x \;\rm M & & + x\;\rm M & & + x\; \rm M \\ \textbf{E} & (0.255782 - x)\; \rm M & & x\; \rm M & & x\; \rm M\end{array}[/tex].

Thus, at equilibrium:

[tex]\displaystyle \frac{[{\rm NH_3}] \cdot [{\rm H^{+}}]}{[{ \rm {NH_4}^{+}}]} = 10^{pK_\text{a}({\rm {NH_4}^{+}})}[/tex].

[tex]\displaystyle \frac{x^2}{0.255782 - x} \approx 10^{-9.25}[/tex]

Solve for [tex]x[/tex]. (Notice that the value of [tex]x\![/tex] is likely to be much smaller than [tex]0.255782[/tex]. Hence, the denominator on the left-hand side [tex](0.255782 - x) \approx 0.255782[/tex].)

[tex]x \approx 1.19929 \times 10^{-5}[/tex].

Hence, the concentration of [tex]\rm H^{+}[/tex] at the equivalence point of this titration would be approximately [tex]1.19929 \times 10^{-5}\; \rm M[/tex].

Hence, the [tex]pH[/tex] at the equivalence point of this titration would be:

[tex]\begin{aligned}pH &= -\log_{10}[{\rm {H}^{+}}] \\ &\approx -\log_{10} \left(1.19929 \times 10^{-5}\right) \approx 4.92\end{aligned}[/tex].

Answer:

5 × 10⁻³ mol

Explanation:

Step 1: Given data

Step 2: Convert "V" to dm³

We will use the conversion factor 1 dm³ = 1000 cm³.

25 cm³ × 1 dm³/1000 cm³ = 0.025 dm³

Step 3: Calculate the moles of HCl (n)

We will use the definition of molarity.

C = n/V

n = C × V

n = 0.2 mol/dm³ × 0.025 dm³ = 5 × 10⁻³ mol