Write the net ionic equation for any precipitation reaction that may be predicted when aqueous solutions of manganese(II) nitrate and sodium hydroxide are combined.

Answer:

1.3 mL

Explanation:

First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:

1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.

It should have 2 significant figures, because 1.2kg has 2 and we are dividing.

The volume of olive oil will be nearly 1300mL or 1.30 L as per the given data.

Volume is a measurement of three-dimensional space that is occupied. It is frequently numerically quantified using SI derived units or various imperial units. The definition of length is linked to the definition of volume.

Volume is, at its most basic, a measure of space. The units liters (L) and milliliters (mL) are used to measure the volume of a liquid, also known as capacity.

This measurement is done with graduated cylinders, beakers, and Erlenmeyer flasks.

Here, it is given that mass of olive oil is 1.2kg.

We know that,

Density of olive oil = 0.917kg/l.

Volume = mass/density

Volume = 1.2/0.917.

Volume = 1.30 lit.

Volume = 1300mL.

Thus, the volume of olive oil will be 1300 mL.

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Answer:

[tex]\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl[/tex].

One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.

Explanation:

Start by finding the formula for each of the compound.

Therefore, the ratio between [tex]\rm K[/tex] atoms and [tex]\rm Br[/tex] atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula [tex]\rm KBr[/tex]. Similarly, the ratio between

The formula for chlorine gas is [tex]\rm Cl_2[/tex], while the formula for bromine gas is [tex]\rm Br_2[/tex].

Write down the equation using these chemical formulas.

[tex]\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl[/tex].

Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both [tex]\rm KBr[/tex] and [tex]\rm KCl[/tex] features two elements each.

Assume that the coefficient of [tex]\rm KCl[/tex] is one. Hence:

[tex]\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl[/tex].

Note that [tex]\rm KBr[/tex] is the only source of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms among the reactants of this reaction.

There would thus be one [tex]\rm K[/tex] atom and one [tex]\rm Br[/tex] atom on the reactant side of the equation.

Because atoms are conserved in a chemical equation, there should be the same number of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms on the product side of the equation.

In this reaction, [tex]\rm Br_2[/tex] is the only product with [tex]\rm Br[/tex] atoms.

One [tex]\rm Br[/tex] atom would correspond to [tex]0.5[/tex] units of [tex]\rm Br_2[/tex].

Similarly, in this reaction, [tex]\rm KCl[/tex] is the only product with [tex]\rm K[/tex] atoms.

One [tex]\rm K[/tex] atom would correspond to one formula unit of [tex]\rm KCl[/tex].

Hence:

[tex]\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].

Similarly, there should be exactly one [tex]\rm Cl[/tex] atom on either side of this equation. The coefficient of [tex]\rm Cl_2[/tex] should thus be [tex]0.5[/tex]. Hence:

[tex]\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].

That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:

[tex]\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl[/tex].

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T}[/tex]

We can solve for the temperature as follows:

[tex]T=\frac{\Delta H_{vap}}{\Delta S_{vap}}[/tex]

Thus, with the proper units, we obtain:

[tex]T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C[/tex]

Hence, answer is approximately 100 °C.

Best regards.

Answer:

Approximately [tex]1.854\; \rm mol\cdot L^{-1}[/tex].

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

Look up the relative atomic mass of [tex]\rm Sr[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

Calculate the formula mass of [tex]\rm Sr(OH)_2[/tex]:

[tex]M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}[/tex].

[tex]M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}[/tex] means that each mole of [tex]\rm Sr(OH)_2[/tex] formula units have a mass of [tex]121.634\; \rm g[/tex].

The question states that there are [tex]10.60\; \rm g[/tex] of [tex]\rm Sr(OH)_2[/tex] in this solution.

How many moles of [tex]\rm Sr(OH)_2[/tex] formula units would that be?

[tex]\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}[/tex].

There are [tex]8.71467\times 10^{-2}\; \rm mol[/tex] of [tex]\rm Sr(OH)_2[/tex] formula units in this [tex]47\; \rm mL[/tex] solution. Convert the unit of volume to liter:

[tex]V = 47\; \rm mL = 0.047\; \rm L[/tex].

The molarity of a solution measures its molar concentration. For this solution:

[tex]\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

(Rounded to four significant figures.)

Answer:

[tex]Kp=0.0386[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]

For which the equilibrium expression is:

[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]

Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:

[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]

And the total pressure:

[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]

Yet it is 748 torr, for which the extent is:

[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]

Therefore, Kp turns out:

[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]

Best regards.

Answer:

Your answer will either be 22.9897 or 22.990 !!

Explanation:

Answer:

too low

Explanation:

If our aim is to recover the naphthalene and measure its mass after separation, then we must not allow any vapour to escape.

Naphthalene is a sublime substance, it can be separated by sublimation. It changes directly from solid to gas. This vapour must be kept securely so that none of it escapes. If part of the naphthalene vapour happens to escape accidentally, then the measured mass of naphthalene will be too low compared to the mass of naphthalene originally present in the mixture.

Answer:

The Nuclear magnetic resonance is the process this technique does not use radiation.

The  ms is an sensitive technology can be a massive number and small sample of the blood.

Explanation:

The Nuclear magnetic resonance we look at the both side of that coin.

The technique provides that fatty acid composition and various including amino acids.

These are contain the complementary these biomarkers, that are suitable for all kinds of studies. there are many types of research:-

(1) A powerful tool metabolic (2) A versatile tool research (3) Quick analysis (4) Low cost analysis.

The MS is an extremely sensitive technology using a very small number of the blood.

(1) Powerful techniques (2) Highly method (3) Large number of metabolites (4)Small sample volume

MS can be fine mapping metabolic pathways to sign analytical strategy.

Answer:

Isomers are molecules that have the same molecular method, however have a unique association of the atoms in space. That excludes any extraordinary preparations which can be sincerely because of the molecule rotating as an entire, or rotating about precise bonds.

Answer:

Explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

.02M

CH₃COOH  = CH₃COO⁻ + H⁺

C                       xC             xC

Ka = xC . xC / C = x² C

1.8 x 10⁻⁵ = x² . .02

x² = 9 x 10⁻⁴

x = 3 x 10⁻²

= .03

concentration of H⁺ = xC = .03 . .02

= 6 x 10⁻⁴ M , volume =  45 x 10⁻³ L

moles of H⁺  = 6 X 10⁻⁴  x 45 x 10⁻³

= 270 x 10⁻⁷ moles

= 2.7 x 10⁻⁵ moles

concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L

moles of Na OH = 2 X 10⁻²  x 35 x 10⁻³

= 70 x 10⁻⁵ moles

=  

NaOH is a strong base so it will dissociate fully .

there will be neutralisation reaction between the two .

Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles

= 67.3 x 10⁻⁵ moles of NaOH

Total volume = 45 + 35 = 80 x 10⁻³

concentration of NaOH after neutralisation.= 67.3  x 10⁻⁵ / 80 x 10⁻³ moles / L

= 8.4125  x 10⁻³ moles / L

OH⁻ = 8.4125  x 10⁻³

H⁺ = 10⁻¹⁴ / 8.4125  x 10⁻³

= 1.1887 x 10⁻¹²

pH = - log (  1.1887 x 10⁻¹² )

= 12 - log 1.1887

= 12 - .075

= 11.925 .

Answer:

0.373 M

Explanation:

The balanced equation for the reaction is given below:

HC2H3O2 + NaOH —> NaC2H3O2 + H2O

From the balanced equation above, the following were obtained:

Mole ratio of the acid, HC2H3O2 (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, we shall write out the data obtained from the question. This include:

Volume of base, NaOH (Vb) = 32.17 mL

Molarity of base, NaOH (Mb) = 0.116 M

Volume of acid, HC2H3O2 (Va) = 10 mL

Molarity of acid, HC2H3O2 (Ma) =..?

The molarity of the acid solution can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 0.116 x 32.17 = 1

Cross multiply

Ma x 10 = 0.116 x 32.17

Divide both side by 10

Ma = (0.116 x 32.17) /10

Ma = 0.373 M

Therefore, the concentration of the acetic acid is 0.373 M.

Answer:

Osmolarity of solution of KCI = 0.40 osmol

Explanation:

Given:

KCL ⇒ K⁺ + Cl⁻

Find:

Osmolarity of solution of KCI

When M = 0.20 M

Computation:

1 mole of KCL = 2 osmol

1 M of KCl = 2 Osmolarity

So,

Osmolarity of solution of KCI = 2 × 0.20

Osmolarity of solution of KCI = 0.40 osmol

Answer:

C

Explanation:

Turning liquid to a solid is like freezing water to ice and requires the water to LOSE (release) heat causing an exothermic reaction.

Answer:

Strontium is smaller

Strontium has the higher ionization energy

Strontium has more valence electrons

Explanation:

It must be understood that both elements belong to the same period i.e the same horizontal band of the periodic table

While Rubidium is an alkali metal(group 1) while Strontium is an alkali earth metal(group 2)

Since they are in the same period, periodic trends would be useful in evaluating their properties

In terms of atomic radius, rubidium is larger meaning it has a bigger atomic size

Generally, across the periodic table, atomic radius is expected to decrease and thus Rubidium which is leftmost is expected to have the higher atomic radius

Since strontium belongs to group 2 of the periodic table, it has 2 valence electrons which is more than the single valence electron that rubidium which is in group 1 has

In terms of ionization energy, the atom with the higher number of valence electrons will have the higher ionization energy which is strontium in this case

Answer:

[tex]pKa=3.70[/tex]

Explanation:

Hello,

In this case, given the information, we can compute the concentration of hydronium given the pH:

[tex]pH=-log([H^+])\\[/tex]

[tex][H^+]=10^{-pH}=10^{-2.18}=6.61x10^{-3}M[/tex]

Next, given the concentration of the acid and due to the fact it is monoprotic, its dissociation should be:

[tex]HA\rightleftharpoons H^++A^-[/tex]

We can write the law of mass action for equilibrium:

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

Thus, due to the stoichiometry, the concentration of hydronium and A⁻ are the same at equilibrium and the concentration of acid is:

[tex][HA]=0.2230M-6.61x10^{-3}M=0.2164M[/tex]

As the concentration of hydronium also equals the reaction extent ([tex]x[/tex]). Thereby, the acid dissociation constant turns out:

[tex]Ka=\frac{(6.61x10^{-3})^2}{0.2164}\\ \\Ka=2.02x10^{-4}[/tex]

And the pKa:

[tex]pKa=-log(Ka)=-log(2.02x10^{-4})\\\\pKa=3.70[/tex]

Regards.

Answer:

Molarity of the sulfuric acid solution is 4.61M

Explanation:

The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:

2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻

That means, 2 moles of base react with 1 mole of sulfuric acid.

If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:

0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄

As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:

0.200 moles H₂SO₄ / 0.0434L = 4.61M

Radiologists are medical doctors that treat injuries using medical imaging (radiology)

Answer:

a person who uses X-rays or other high-energy radiation, especially a doctor specializing in radiology.

Explanation:

Answer:

An addiction could occur, maybe an overdose?, this could lead to death and maybe you would do unreasonable things which could get you fined or arrested.

Explanation:

Answer:

Hope this helped,

Kavitha

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

The given question is incomplete. The complete question is :

Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?​

a) [tex]O^{2-}[/tex]

b)  [tex]F^{-}[/tex]

c)  [tex]N^{3-}[/tex]

d)  [tex]S^{2-}[/tex]

Answer: b)  [tex]F^{-}[/tex]

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

Here potassium is having an oxidation state of +1 called as  cation and thus is an anion must have an oxidation state of -1 if they have to combine in 1: 1 ratio to  give neutral ionic compound.

Thus the anion has to be [tex]F^-[/tex] which combines with [tex]K^+[/tex] in 1: 1 ratio to give [tex]KF[/tex]

Answer:

Explanation:

a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .  

C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O

O is provided by KMnO₄

b ) In this reaction isophthalic acid is formed .

C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂

c)

4-Propyl-3-t-butyltoluene

In this oxidation , three side chains of ring  are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .

The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )

Answer:

The correct option is d

Explanation:

Nuclide is synonymous with groups of electrons or protons, that is, a nuclide is the grouping of nucleons.

Answer:

(a) The diode voltage,  [tex]V_D =[/tex]  0.776 V

(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A

Explanation:

Given;

saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A

nonideality factor, n = 1.05

(a) the diode voltage

Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A

Diode voltage is calculated as;

[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]

Where;

[tex]V_T[/tex] is thermal voltage at 25°C = 0.025

[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]

b) the diode current for VD = 0.1 mV

[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]

Answer:

Hey there!

That would be the alkaline earth metals.

Hope this helps :)

Answer: alkaline earth metals

Explanation:

Answer:

Do not conduct electricity as solids.

Explanation:

Hello,

In this case, we should remember that salts are formed when an acid and base react in order to yield the salt and water due to the ions exchange during neutralization chemical reactions. For instance, when hydrochloric acid  (acid) reacts with potassium hydroxide (base), sodium chloride (salt) and water are yielded via:

[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]

Moreover, it is widely known that salts are formed by electrovalent/ionic bonds which involves electron transfer so the metallic atom becomes positively charged (cation) whereas the non-metallic atom becomes negatively charged (anion) once the electrons are received so it can conduct electricity when dissolved in water yet not when solid since electron transfer is facilitated by the aqueous media, otherwise, ions remain together. Thereby, answer is do not conduct electricity as solids.

Regards.

Answer:

c

Explanation:

Answer:

6

Explanation:

Alkanes undergo substitution reaction so the number of replacement reaction hydrogen is 6

Answer:

37%

Explanation:

From the question, the equation goes does.

C2H6+ (1-x)+a(O2+3.76N2)=bC02 + cH2O + axO2 + 3.76dN2.

Mair=Mair/Rin

( MN)O2 + (MN)N2÷ (MN)O2 + (MN)N2 +(MN)C2H6.

33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1

= 176/176+8

X= 0.37

0.37 × 100

X= 37%

Answer:

Water

Explanation:

Solid potassium hydrogen tartrates (KHT) is soluble in water. This is especially at room temperature.

The solvent for KHT is water.