what is the molality of a solution that has 4 mol of KCI in 0.800 kg of water?
help me pleaseeee​

Answer:

For gases such as hydrogen, oxygen, nitrogen, helium, or neon, deviations from the ideal gas law are less than 0.1 percent at room temperature and atmospheric pressure. Other gases, such as carbon dioxide or ammonia, have stronger intermolecular forces and consequently greater deviation from ideality.

Explanation:

Answer: The instantaneous rate of formation of HBr at 50 s is [tex]1.4\times 10^{-2}M/s[/tex]

Explanation:

From the graph,

Initial rate of the [tex]Br_2[/tex] = 1.0 M

Time when the concentration of [tex]Br_2[/tex] is 0.5 M (half the concentration ) = 60 sec

For first order reaction:

Calculating rate constant for first order reaction using half life:

[tex]t_{1/2}=\frac{0.693}{k}[/tex] .....(1)

[tex]t_{1/2}[/tex] = half life period = 60 s

k = rate constant = ?

Putting values in equation 1:

[tex]k=\frac{0.693}{60s}\\\\k=0.01155s^{-1}[/tex]

For the given chemical reaction:

[tex]H_2(g)+Br_2(g)\rightarrow 2HBr(g)[/tex]

Rate of the reaction = [tex]-\frac{\Delta [Br_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}[/tex]

Negative sign represents the disappearance of the reactants

From the above expression:

[tex]k[Br_2]=-\frac{\Delta [Br_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}[/tex]

At 50 seconds, [tex][Br_2]=0.6 M[/tex]

Plugging values in above expression, we get:

[tex]\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}=0.01155\times 0.6\\\\\frac{\Delta [HBr]}{\Delta t}=2\times 0.01155\times 0.6=0.01386=1.4\times 10^{-2}M/s[/tex]

Hence, the instantaneous rate of formation of HBr at 50 s is [tex]1.4\times 10^{-2}M/s[/tex]

Answer:

3.24 mol/L

Explanation:

Given that:

mass of Boron triiodide = 6664 grams

molar mass of BI_3 = 391.52 g/mol

Recall that:

number of moles = mass/molar mass

∴

number of moles = 6664 g /391.52 g/mol

number of moles = 17.02 mol

Also;

Molarity = moles for solute/liter for solution

= 17.02 mol/5.25 L

= 3.24 mol/L

Answer:

14 mol e⁻

Explanation:

Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese

8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)

Step 2: Calculate the moles corresponding to 110 g of manganese

The molar mass of Mn is 55 g/mol.

110 g × 1 mol/55 g = 2 mol

Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn

According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.

2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻

Answer

• Silver

• copper

Answer:

Mass of mercury produced = 2.00 g

Explanation:

At room temperature and pressure, the temperature T = 25° C or 298.15 K while the pressure = 1 atmosphere or 760 mmHg.

Mass of oxide of mercury decomposed at room temperature and pressure = 2.16 g

Volume of oxygen produced = 120 cm³ or 0.12 dm³

One mole of any gas has a volume of 24 dm³ at room temperature and pressure.

Therefore, number of moles of oxygen produced = 0.12 dm³ / 24 dm³/mol = 0.005 moles

Mass of oxygen produced = number of moles × molar mass

Molar mass of oxygen = 32 g/mol

Mass of oxygen produced = 0.005 moles × 32 g/mol = 0.16 g

Thus, mass of mercury produced = mass of mercury oxide decomposed - mass of oxygen produced

Mass of mercury produced = 2.16 g - 0.16 g

Mass of mercury produced = 2.00 g

Explanation:

57.3ml

we use Charles's law

to solve the question

Answer:

[tex]\boxed {\boxed {\sf 225 \ mL}}[/tex]

Explanation:

The temperature and volume of the gas are changing, so we use Charles's Law. This states the temperature of a gas is directly proportional to the volume of a gas. The formula is:

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

The original volume is unknown. The new volume is 75 milliliters.

The gas is cooled from 150 Kelvin to 50 Kelvin, so the original temperature is 150 K and the new temperature is 50 K.

We know that:

Substitute the values into the formula.

[tex]\frac {V_1}{150 \ K}=\frac{ 75 \ mL}{50 \ K}[/tex]

Since we are solving for the original volume, we must isolate the variable V₁.

It is being divided by 150 K. The inverse of division is multiplication, so we multiply both sides by 150 K.

[tex]150 \ K *\frac {V_1}{150 \ K}=\frac{ 75 \ mL}{50 \ K}* 150 \ K[/tex]

[tex]V_1=\frac{ 75 \ mL}{50 \ K}* 150 \ K[/tex]

The units of Kelvin (K) cancel.

[tex]V_1= \frac{ 75 \ mL}{50 }* 150[/tex]

[tex]V_1=1.5 * 150 \ mL[/tex]

[tex]V_1= 225 \ mL[/tex]

The original volume is 225 milliliters.

Answer: This is the hue of mercury if you're looking for it. Mercury reacts with sulphur to generate a crimson hue. In the past, they always used a thermometer.

Explanation:  However, due of its toxicity, individuals have switched to using alcohol in glass thermometers, which work similarly to mercury thermometers.

hope this helped best of luck mate! :) if this helped make sure to mark me Brainliest!

Answer:

c. NO2

b. NH3

d. SO4 ^2 -

a. HPO4 ^ 2 -

Explanation:

Acid is a compound which ionizes to produce hydrogen ions. The Ph value for acid is below 7. Base is a compound which ionizes to produce hydroxide ions.  The Ph value for base is above 7. The Conjugate base accepts a proton or releases a hydrogen ion.

Answer:

[tex]V_{Na_2S}=368mL[/tex]

[tex]m_{Al_2S_3}=67.3gAl_2S_3[/tex]

Explanation:

Hello there!

In this case, according to the given information, it is possible to realize that the only way for us to calculate the required volume of sodium sulfide, is by calculating the moles of this substance consumed 163 mL of 2.75 mol/L aluminum sulfate by using the definition of molar concentration and the 1:3 mole ratio between these two:

[tex]n_{Na_2S}=0.163L*2.75\frac{molAl_2(SO_4)_3}{L}*\frac{3molNa_2S}{1molAl_2(SO_4)_3} =1.34molNa_2S[/tex]

Now, we divide these moles by the molar concentration of sodium sulfide to obtain the required volume:

[tex]V_{Na_2S}=\frac{1.34molNa_2S}{3.65mol/L} =0.368L=368mL[/tex]

For the last part, we now use the 1:1 mole ratio of aluminum sulfate to aluminum sulfide and the molar mass of the latter (150.158 g/mol) in order to calculate the required mass:

[tex]m_{Al_2S_3}=0.163L*2.75\frac{molAl_2(SO_4)_3}{L}*\frac{1molAl_2S_3}{1molAl_2(SO_4)_3} *\frac{150.158gAl_2S_3}{1molAl_2S_3} \\\\m_{Al_2S_3}=67.3gAl_2S_3[/tex]

Regards!

Answer: The number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].

Explanation:

The reaction equation is as follows.

[tex]FeCl_{3}(aq) + 3NaOH(aq) \rightarrow Fe(OH)_{3}(s) + 3NaCl(aq)[/tex]

Therefore, moles of [tex]Fe(OH)_{3}[/tex] are calculated as follows.

Moles = Molarity of [tex]Fe(OH)_{3}[/tex]  [tex]\times[/tex] Volume (in L)

= 0.654 M [tex]\times[/tex] 0.197 L  

= 0.128 mol

Now, according to the given balanced equation 1 mole of [tex]FeCl_{3}(aq)[/tex] reacts with 3 moles of NaOH(aq). Hence, moles of [tex]Fe(OH)_{3}[/tex]  reacted are calculated as follows.

3 [tex]\times[/tex] 0.128 mol = 0.384 moles of NaOH

As moles of NaOH present are as follows.

Moles of NaOH = Molarity of NaOH [tex]\times[/tex] Volume (in L)

0.384 mol = 0.587 M [tex]\times[/tex] Volume (in L)

Volume (in L) = 0.654 L (1 L = 1000 mL) = 654 mL

Thus, we can conclude that the number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].

Answer:

false

Explanation:

energy does not return to the sun, it returns to the plants or producers. if energy were to return to the sun, it would have to travel though space.

Answer:

Density

Explanation:

The mass of an element is the average weight that the isotope of the particular element contains. Its characteristic indicates the amount of substance present in an element.

However, the volume of an element on the other hand is the mole of an element that is contained in a room temperature.

The relation joining both the mass and volume of an element is density.

This is because density showcase the relationship between the mass of an element to space in occupies in terms of volume.

It is given by the formula:

Density = mass/volume

Answer:

Which language is this???

Answer:

True

Explanation:

Your welcome! :) Good luck!

Answer:

The concentration of the acid, HCl is 0.013 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nₐ) = 2

The mole ratio of base, Ca(OH)₂ (n₆) = 1

Finally, we shall determine the molarity of the HCl. This can be obtained as follow:

Volume of acid, HCl (Vₐ) = 250 mL

Molarity of base, Ca(OH)₂ (M₆) = 0.118 M

Volume of base, Ca(OH)₂ (V₆) = 13.7 mL

Molarity of acid, HCl (Mₐ) =?

MₐVₐ / M₆V₆ = nₐ/n₆

Mₐ × 250 / 0.118 × 13.7 = 2/1

Mₐ × 250 / 1.6166 = 2

Cross multiply

Mₐ × 250 = 1.6166 × 2

Mₐ × 250 = 3.2332

Divide both side by side 250

Mₐ = 3.2332 / 250

Mₐ = 0.013 M

Thus, the concentration of the acid, HCl is 0.013 M

Answer:

0.016 M

Explanation:

Molarity refers to the molar concentration of a solution and it can be calculated using the formula below:

Molarity (M) = number of moles (n) ÷ volume (V)

According to this question, the mass of KCl was given to be 0.72 grams and the volume of water as 600 mL.

Using mole = mass/molar mass to convert mass of KCl to moles

Molar mass of KCl = 39 + 35.5 = 74.5g/mol

mole = 0.72g ÷ 74.5g/mol

mole = 0.00966mol

Volume of water = 600mL = 600/1000 = 0.600L

Molarity, M = 0.00966 ÷ 0.600

Molarity of KCl solution = 0.016 M

Answer:

1.15 moles of excess reactant will remain after precipitation is complete.

Explanation:

The balanced reaction is:

2 AgNO₃ (aq) + Na₂SO₄ (aq) → Ag₂SO₄ (s) + 2 NaNO₃ (aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Then you can apply the following rule of three:: if by stoichiometry 2 moles of AgNO₃ reacts with 1 mole of Na₂SO₄, 3.80 moles of AgNO₃ reacts with how much moles of Na₂SO₄?

[tex]amount of moles of Na_{2}SO_{4} =\frac{1mole of Na_{2}SO_{4} * 3.80 moles of AgNO_{3} }{2 mols of AgNO_{3} }[/tex]

amount of moles of Na₂SO₄= 1.9 moles

But 3.05 moles of Na₂SO₄ are available. Since you have more moles than you need to react with 3.80 moles of AgNO₃, Na₂SO₄ will be the excess reagent.

To calculate the amount of excess reagent that will remain, you must make the difference between the amount you initially have and the amount that reacts:

3.05 moles - 1.9 moles= 1.15 moles

1.15 moles of excess reactant will remain after precipitation is complete.

Answer:

something that can be maintained over a period of time

Answer:

a balance between meeting today's needs.......

Explanation:

Answer:

5 and the rest are all set to the same date on your list as the other one to get you a list on for a your special first year week and with a special holiday party holiday

Explanation:

Sorry desperate for points

Answer:

F2 (g)

Explanation:

Edg 2021

Answer:

F2 g

Explanation:

Answer:

Correct answer would be Option 2, Chemical Energy

Hope this helps!

Solution :

The equation is :

[tex]$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$[/tex]

The number of the moles of HA os 0.00285, and the volume is 25 mL.

15 mL of the 0.0950 M NaOH is added.

The total volume of a solution is V = 25 mL + 15  mL = 40 mL

The pH of the solution is 6.50

Calculating the [tex]K_a[/tex] of HA

[tex]$HA(aq) \rightleftharpoons A^-(aq)+H^+$[/tex]

[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]

Let s calculate the concentration of HA and NaOH

[tex]$[HA] = \frac{^nH_A}{V}$[/tex]

        [tex]$=\frac{0.00285 \ mol}{0.04 \ L}$[/tex]

       = 0.07125 M

[tex]$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$[/tex]

            [tex]$=\frac{0.001425 mol}{0.04L}$[/tex]

           = 0.0356 M

                                      [tex]$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$[/tex]

Initial conc. (M)            0.07125 M       0.0356 M            0 M

Change in conc. (M)   -0.0356 M       -0.0356 M        + 0.0356 M

Equilibrium conc. (M)   0.03565 M        0 M                0.0356 M

Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M

0.0356 M of NaA dissociates completely into 0.0356 M [tex]Na^+[/tex] and 0.0356 M [tex]A^-[/tex]

Now for [tex][H^+][/tex]

[tex]$[H^+] = 10^{-pH}$[/tex]

       [tex]$=10^{-6.5}$[/tex]

       [tex]$=3.16 \times 10^{-7}$[/tex]

Calculating the value of [tex]K_a[/tex],

[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]

     [tex]$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$[/tex]

     [tex]$=3.16\times 10^{-7}$[/tex]

Therefore the the value of [tex]K_a[/tex] for the unknown acid is [tex]$3.16\times 10^{-7}$[/tex].

Protons: charge: +1 // mass: 1 // location: nucleus

Neutrons: charge: 0// mass: 1 // location: nucleus

Electrons: charge: -1// mass: 0// location: orbitals

Answer: Rank in order the strongest to the weakest acid is [tex]CCl_{3}COOH[/tex] >  [tex]CBr_{3}COOH[/tex] >  [tex]CH_{3}COOH[/tex].

Explanation:

More readily a substance is able to donate a hydrogen ion more will be its acidic strength. Hence, stronger will be the acid.

More is the electronegativity of atoms attached to the acid more easily it will donate a proton. Hence, more will be its acidic strength.

Chlorine is more electronegative in nature as compared to bromine. So,

[tex]CCl_{3}COOH[/tex] is more acidic than [tex]CBr_{3}COOH[/tex].

Since there is no electronegative group attached to [tex]CH_{3}COOH[/tex] so it is least acidic than [tex]CCl_{3}COOH[/tex] and [tex]CBr_{3}COOH[/tex].

Thus, we can conclude that rank in order the strongest to the weakest acid is [tex]CCl_{3}COOH[/tex] >  [tex]CBr_{3}COOH[/tex] >  [tex]CH_{3}COOH[/tex].

Answer:

46.67 g

Explanation:

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

Change in temperature (ΔT) = 23.8 °C

Heat (Q) = 261 J

Specific heat capacity (C) of silver = 0.235 J/gºC

Mass of silver (M) =?

The mass of the sample of silver can be obtained as follow:

Q = MCΔT

261 = M × 0.235 × 23.8

261 = M × 5.593

Divide both side by 5.593

M = 261 / 5.593

M = 46.67 g

Thus, the mass of the sample of silver is 46.67 g

Answer:

D

Explanation:

because product are at the top and reactant are at the bottom also it to the power of the moles in front e.g 2NO it will be to the power of 2 in this case.

hope it make sense :)

Answer:

b) warming up a) wavelength a) blank c) sample

Explanation:

To run a spectrophotometry experiment, begin by warming up the spectrophotometer and preparing the samples. It is important that the equipment is warmed up for at least 30 minutes before starting the measurements.

Be sure to select the correct wavelength, then run a measurement on the blank solution. The selected wavelength depends on the analyte of interest. The black solution contains the same matrix but it doesn´t contain the analyte.

Follow up by running measurements on sample solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples. The samples are those of unknown concentration that we want to determine.

To run a spectrophotometry experiment, begin by cleaning the spectrophotometer and preparing the samples. Be sure to select the correct wavelength, then run a measurement on the sample solution. Follow up by running measurements on aqueous solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples.

Spectrophotometry is a technique used to measure the interaction of light with matter, specifically the absorption, transmission, or reflection of light by a sample. It involves the use of a spectrophotometer, an instrument that measures the intensity of light as a function of its wavelength or frequency.

In spectrophotometry, a sample is exposed to light of a specific wavelength or a range of wavelengths. The sample may absorb certain wavelengths of light, which can be detected and measured by the spectrophotometer. The amount of light absorbed is related to the concentration of the analyte in the sample, allowing for quantitative analysis.

Learn more about Spectrophotometry, here:

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