Describe how to test your unknown salt mixture for the presence of
Na3PO4.12H2O.​

Answers

Answer 1

Na3PO4*12H2O + BaCl2*2H2O = Ba3(PO4)2 + NaCl + H2O

add barium chloride to your Na3PO4.12H2O a white precipitate of Ba3(PO4)2 will be formed wrt salt(NaCl) and water(H20) if Na3PO4.12H2O. will be there.


Related Questions

What is the [OH-] in a solution if the [H*] = 1.2 x 10-3 M?

Answers

We know that [OH⁻] * [H⁺] = 10⁻¹⁴

plugging the value of [H⁺]

[OH⁻] * 1.2 * 10⁻³ = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ * (10³/1.2)

[OH⁻] = 833.3 * 10⁻¹⁴

[OH⁻] = 8.33 * 10⁻¹²

The specific heat of palladium is 0.239 J/g°C. How much heat (q) is released when a 10.0 g piece of platinum cools from 100.0°C to 50.0°C?

Answers

Answer:

119.5 J

Explanation:

First we calculate the temperature difference:

ΔT = 100 °C - 50 °C = 50 °C

Then we can calculate the heat released by using the following formula:

q = Cp * ΔT * m

Where q is the heat, Cp is the specific heat, ΔT is the temperature difference and m is the mass.

We input the data:

q = 0.239 J/g°C * 50 °C * 10.0 gq = 119.5 J

How many Grams of NO is produced if 12g of O2 is combined with excess ammonia?

Answers

Answer:

9g

Explanation:

moles O2 = mass / Mr = 12 / 2(16.0) = 0.375

ratio O2 : NO = 5:4

moles NO produced = 0.375 * 4/5 = 0.3

mass NO = Mr * mol = (14.0+16.0) * 0.3 = 9g

The Grams of NO is produced if 12g of O2 is combined with excess ammonia is 80 gram.

What is NO?

The full form of NO is nitrogen oxide Nitrogen belong to the 15th group of the periodic table with molar mass of 14 and oxygen belong to the 16th group of the periodic table.

In the given their are 4 molecules of each nitrogen and oxygen and mass of O2 is given whereas, mass of nitrogen for 4 molecule will be 56 gram.

Grams of NO = 14 × 4 + 12 × 2

Grams of NO = 80 grams.

Therefore, the mass of NO for the given reaction is combined with excess ammonia is 80 gram.

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explain the trend in boling point of alcohols in their homologous series​

Answers

Explanation:

The boiling point of alcohols also increase as the length of hydrocarbon chain increases

The boiling point of alcohols also increase as the length of hydrocarbon chain increases.

Why boiling points of alcohols increases as the length of hydrocarbon chain increasesSince, boiling point depends upon type of bonds present in the molecules.If molecules have hydrogen bonding then boiling point is higher than others because hydrogen bond is stronger bond.So, higher energy or temperature required to break these bond or overcomes the inter molecular force of attraction.Since, alcohol contains hydrogen bonding. So, its boiling points are higher as compare to other compounds.Since, boiling points is also directly proportional to molar mass of molecules. So, molecule have higher molar mass has comparatively high boiling point.Example: An alcohols with lower hydrocarbon chain has lower boiling points as compare to alcohols with higher  hydrocarbon chain.

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A single ___ bond is made when two atoms share a pair of ____

Answers

Answer:

covalent

valence electrons

Explanation:

The attraction between two atoms that share a pair of valence electrons is known as a covalent bond. The nuclei of both atoms are drawn to the shared electrons. This results in a molecule with two or more atoms. Covalent bonds are formed solely between nonmetal atoms.

Between atoms of the same element or between atoms of different elements, covalent bonds can form. A new substance termed a covalent compound is formed when atoms of various elements create covalent bonds.

A sample of aluminum absorbed 9.86 J of heat and its temperature increased from 23.2 and 30.5 degrees * C . What is the mass of the aluminum? Th specific heat of aluminum is 0.902 J/g^ C . Round your answer to 2 significant figures. Do not include units in your answer. *

Answers

Explanation:

H=mc×∆©

9.86=m×0.902×(30.5-23.2)

m=1.5

Explanation:

The specific heat of a substance is the amount of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. The formula for calculating the heat absorbed or released by a substance is `q = mcΔT`, where `q` is the heat absorbed or released, `m` is the mass of the substance, `c` is the specific heat of the substance, and `ΔT` is the change in temperature.

In this case, we can use this formula to solve for the mass of the aluminum sample. We know that `q = 9.86 J`, `c = 0.902 J/g°C`, and `ΔT = 30.5°C - 23.2°C = 7.3°C`. Plugging these values into the formula, we get:

`9.86 J = m * 0.902 J/g°C * 7.3°C`

Solving for `m`, we find that the mass of the aluminum sample is approximately `1.5 g`, rounded to 2 significant figures.

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A gas stp is 4.38 L how many moles are there ?

Answers

Answer:

There are 0.195 moles.

Explanation:

1 mol of any gas under STP (meaning 1 atm and 273 K) will occupy 22.4 liters.

With the information above in mind, we can calculate how many moles will occupy 4.38 liters:

4.38 L * 1 mol / 22.4 L = 0.195 mol

In 4.38 liters of a gas at STP there will be 0.195 moles.

问题2
15
The reaction causes the temperature of its surroundings to decrease is
therefore heat is
by the reaction.
and
Select the two words that best complete this sentence.
a. endothermic, absorbed
b. exothermic, released
c. exothermic, absorbed
d. endothermic, released

Answers

Answer:

a endothermic, absorbed

What mass of oxygen will react with 2.64 g of magnesium?

2Mg(s) + O2(g) → MgO(s)

Answers

Answer:

[tex](24 \times 2) \: g \: of \: magnesium \: reacts \: with \: (16 \times 2) \: g \: of \: oxygen \\ 2.64 \: g \: of \: magnesium \: will \: react \: with \: ( \frac{2.64 \times 16 \times 2}{24 \times 2} ) \: g \\ = 1.76 \: g \: of \: oxygen[/tex]

write 5 acid and 5 akalis

Answers

Answer:

5 Acids

Chemical

- Hydrochloric acid in gastric juice

- Sulphuric acid

- Nitric acid

- Uric acid

- Acetic acid

Household

- Vinegar

- Lemon

- Milk

- Batteries

- Soft drinks

5 Alkali

Household

- Dishwashwer soaps

- Detergents

- Toothpaste

- Oven cleaner

- Alkaline batteries

- Sodium bicarbonate

Chemicals

- Sodium hydroxide or lye

- Calcium carbonate (limestone)

- Ammonium hydroxide

- Calcium hydroxide

cloruro de litio mas sodio​

Answers

Answer:

El cloruro de litio se utiliza para fabricar litio metálico. El cloruro de litio se funde y electroliza. Esto produce litio metálico líquido.

Explanation:

sowwy si mi español es malo

Answer:

give him brainliest

Explanation:

Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

Answers

Answer: There are [tex]16.14 \times 10^{23}[/tex] atoms of hydrogen are present in 40g of urea, [tex](NH_{2})_{2}CO[/tex].

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol[/tex]

According to the mole concept, 1 mole of every substance contains [tex]6.022 \times 10^{23}[/tex] atoms.

So, the number of atoms present in 0.67 moles are as follows.

[tex]0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms[/tex]

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

[tex]4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms[/tex]

Thus, we can conclude that there are [tex]16.14 \times 10^{23}[/tex] atoms of hydrogen are present in 40g of urea, [tex](NH_{2})_{2}CO[/tex].

What is the main function of a nucleic acid?
A. It allows polysaccharides to form.
OB. It stores lipids inside the proteins.
C. It contains instructions for making proteins.
OD. It is a catalyst in chemical reactions.

Answers

The main function of nucleic acid is It stores lipids inside the proteins.

Nucleic acids

Nucleic acids are large biomolecules that play essential roles in all cells and viruses. A major function of nucleic acids involves the storage and expression of genomic information. Deoxyribonucleic acid, or DNA, encodes the information cells need to make proteins. A related type of nucleic acid, called ribonucleic acid (RNA), comes in different molecular forms that play multiple cellular roles, including protein synthesis. 

Nucleic acids are made of nitrogen-containing bases, phosphate groups, and sugar molecules. Each type of nucleic acid has a distinctive structure and plays a different role in our cells. Researchers who first explored molecules inside the nucleus of cells found a peculiar compound that was not a protein or a lipid or a carbohydrate. It was new.  The discovery of this molecule — nuclein, which upon further understanding became nucleic acid — set in motion the eventual discovery of DNA.

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What mass of HF is produced according to the given equation when 2.397 grams of each reactants are combined?

CaF2 + H2SO4 --> CaSO4 + 2HF​

Answers

+HF is the answer

Explanation:

How many moles are 2.60 * 10 ^ 27 atoms of Silicon?

Answers

Answer:

~4,317.5 moles of Silicon

Explanation:

Using Avogadro's contant we know that:

1mole = 6.022 x 10^23 atoms

So,to calculate the number of moles in 2.60 x 10^27 atoms of Silicon:

=(2.60 x 10^27 x 1)/(6.022 x 10^23)

~4,317.5 moles of Silicon

Hope it helps:)

If 1.00 mol of an ideal monatomic gas initially at 74 K absorbs 100 J of thermal energy, what is the final temperature

Answers

Answer:

T = 82 K

Explanation:

The computation of the final temperature is shown below;

Given that

T_0 denotes the initial temperature of the gas i.e. 74 K

T denotes the final temperature of the gas = ?

n denotes number of moles of monoatomic gas i.e. 1.00 mol

R denotes universal gas constant = 8.314

c denotes the heat capacity at constant volume i.e.

= (1.5) R = (1.5) (8.314)

= 12.5

Q denotes the Amount of heat absorbed i.e 100 J

We know that

Amount of heat absorbed is provided as

Q = n c (T - T_0)

100 = (1) (12.5) (T - 74)

T = 82 K

HELP PLZ
Calculate the percent composition by mass of iron in Fe(NO3)3

Answers

hey ik this isn’t a big help but there is a Chem app you can use to get your answers faster rather than waiting for someone to respond. anyways the app is called, ChemCalculator.

How much energy is released when 6.0 g of water is condensed from water
vapor?
A. 6.0 g x 1 mol/18.02 g x 4.186 kJ/mol
B. 6.0 g 1 mol/18.02 g 6.03 kJ/mol
O C. 6.0 g x 1 mol/18.02 g * (-285.83 kJ/mol)
O D. 6.0 g x 1 mol/18.02 g x 40.65 kJ/mol

Answers

Answer: 6.0g x 1 mol/18.02g x 40,65 kJ/mol which is D

Explanation: Just did

The water vapors change from vapor into water in condensation. The energy released by 6 gm of water is 6.0 g x 1 mol/18.02 g x 40.65 kJ/mol. Thus, option D is correct.

What is the heat of vapourization?

The heat of vapourization is the latent heat or enthalpy needed by the liquid to get converted into the vapor or the gaseous phase of the matter. It can be used to determine the energy released or absorbed by the substance.

The heat of vapourization of water is 40.65 kJ/mol, and its molar mass is 18.02 moles.

So, the energy released for the water when it is condensed into vapors will be,

6.0 g x 1 mol/18.02 g x 40.65 kJ/mol

Therefore, the heat of vapourization determines the amount of energy released.

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1. For the reaction 3A — C, the initial concentration of A was 0.2 M, and the reaction rate was
1.0 M/s. When [A] was doubled, the reaction rate increased to 4.0 M/s. Determine the rate
law for the reaction.

Answers

Answer:

[tex]r=25M^{-1}s^{-1}[A]^2[/tex]

Explanation:

Hello there!

In this case, according to the given information for this chemical reaction, it is possible for us to set up the following general rate law and the ratio of the initial and the final (doubled concentration) condition:

[tex]r=k[A]^n\\\\\frac{r_1}{r_2} =\frac{k[A]_1^n}{k[A]_2^n}[/tex]

Next, we plug in the given concentrations of A, 0.2M and 0.4 M, the rates, 1.0 M/s and 4.0 M/s and cancel out the rate constants as they are the same, in order to obtain the following:

[tex]\frac{1.0}{4.0} =\frac{0.2^n}{0.4^n}\\\\0.25=0.5^n\\\\n=\frac{ln(0.25)}{ln(0.5)} \\\\n=2[/tex]

Which means this reaction is second-order with respect to A. Finally, we calculate the rate constant by using n, [A] and r, to obtain:

[tex]k=\frac{r}{[A]^n} =\frac{1.0M/s}{(0.2M)^2}\\\\k=25M^{-1}s^{-1}[/tex]

Thus, the rate law turns out to be:

[tex]r=25M^{-1}s^{-1}[A]^2[/tex]

Regards!

What are quarks?
A. Particles that bind gluons together within the nucleus
B. Radioactive material that is emitted from a nucleus
C. Subatomic particles that make up protons and neutrons.
D. Fundamental forces acting between two pieces of matter

Answers

Answer:

D. Fundamental forces acting between two pieces of matter

Answer:

D. Fundamental forces acting between two pieces of matter

What transition energy corresponds to an absorption line at 460 nm?
A. 6.52 x 10-19 J
B. 4.32 x 10-19 J
C. 4.45 x 10-19 J

D. 2.31 x 10-19 J

Answers

The answer is b I think I hope it right

Answer:

B. 4.32 x 10-19 J is correct via a p e x

Explanation:

The compound aluminum nitride () is a compound semiconductor having mixed ionic and covalent bonding. The electronegativities for and are 1.5 and 3.0 respectively. Calculate the fraction of the bonding that is ionic.

Answers

Answer:

Fraction ionic = 0.43

Explanation:

To solve this question, we must, as first, find the fraction of the bond that is covalent using the equation:

Fraction Covalent = exp (-0.25*(Enitride - EAl)²)

Fraction covalent = exp (-0.25*(3.0- 1.5)²)

Fraction covalent = exp (-0.25*2.25)

Fraction covalent = exp (-0.5625)

Fraction covalent = 0.57

As:

1 = Fraction Covalent + Fraction Ionic:

Fraction ionic = 1 - 0.57

Fraction ionic = 0.43

How many moles of MgCO3 are present in 252.939 grams of MgCO3?

A. 2
B. 3
C.5
D.6

Answers

Answer:

Hello, I was doing this and from my caluclations it should be :

Explanation:

3 Moles.

The three moles of MgCO3 are present in 252.939 grams of MgCO3.

So, option B is correct one .

What is molar mass?

The molar mass of substance is equal to mass of one mole of  that substance.

Example: Molar mass of MgCO3 is equal to mass of one mole of MgCO3.

Molar mass of Mg = 24.305 u

Molar mass of C = 12.011 u

Molar mass of O = 15.999 u

Molar mass of MgCO3 = Molar mass of Mg + Molar mass of C + Molar mass of O*3

Molar mass of MgCO3 = 24.305 + 12.011  +  15.999 * 3

Molar mass of MgCO3 = 84.3139 gram

Molar mass of  MgCO3 contain  84.3139 gram

So, 252.939 grams of MgCO3 = 252.939 grams/84.3139 gram

252.939 grams of MgCO3 = 2.999 moles

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The area of a telescope lens is 6507 x 10^3 mm^2. What is the area in square feet, enter your answer in scientific notation. If it takes a technician 51.6 s to polish 1.68 x 10^2 mm^2 how long does it take her to polish the entire lens ?

Answers

Answer:  the area of the telescope lens is \textit{0,08507 ft}

Explanation: Happy I could help!

Which waves are blocked by the atmosphere? A. gamma rays B. visible light C. radio waves D. infrared waves​

Answers

Answer: look at the explanation and try to work it

Explanation: in contrast, our atmosphere blocks most ultraviolet light (UV) and all X-rays and gamma-rays from reaching the surface of Earth. Because of this, astronomers can only study these kinds of light using detectors mounted on weather balloons, in rockets, or in Earth-orbiting satellites.

PLEASE HELP ME!!!!!!!

Answers

Answer:

The heat capacity of the metal underneath the gold is 0.431 J/g°C

Explanation:

Using the formula as outlined in the image:

Q = m × c × ∆T

Where;

Q = amount of heat energy (J)

m = mass of substance (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

According to the information in this question;

Q = 503.9J

m = 23.02g

c = ?

∆T = 74°C - 23.2°C = 50.8°C

Using Q = m × c × ∆T

c = Q ÷ m∆T

c = 503.9 ÷ (23.02 × 50.8)

c = 503.9 ÷ 1169.42

c = 0.431 J/g°C

From the above heat capacity of the metal underneath the gold, it is obvious that the metal is not pure gold (c = 0.129J/g°C)

Using the periodic table as a reference, which pair of elements are nonmetals?

A. oxygen and sulfur

B. cobalt and zinc

C. mercury and lead

D. sodium and iodine

Answers

Answer:

A). oxygen and sulphur

Non metals are on the right side of the periodic table they are generally in the gaesous form there are as many as 20 non metals in the periodic group ( including halogens and noble gas). non metal are electron withdrawning groups and they generally forms anion they are poor conductors of heat and electricity

You need to prepare 1 L of the citric acid/citrate buffer. You have chosen to use Method 1 (see lab presentation). Calculate the pH of a solution prepared by mixing 300 mL of 0.45 M citric acid acid and 100 mL of 0.65 M NaOH plus water to a final volume of one liter. The Ka of citric acid is 7.24 x 10-4.

Answers

Answer:

3.11 is the pH of the buffer

Explanation:

The pH of a buffer is obtained using H-H equation:

pH = pKa + log [Conjugate base] / [Weak acid]

Where pH is the pH of the buffer, pKa = -log Ka = 3.14 for the citric buffer and [] could be taken as the moles of each species.

The citric acid,HX (Weak acid), reacts with NaOH to produce sodium citrate, NaX (weak base) and water:

HX + NaOH → H2O + NaX

That means the moles of NaOH added = Moles of sodium citrate produced

And the resulitng moles of HX = Initial moles - Moles NaOH added

Moles HX and NaX:

Moles NaOH = 0.100L * (0.65mol / L) = 0.065 moles NaOH = Moles NaX

Moles HX = 0.300L * (0.45mol / L) = 0.135 moles HX - 0.065 moles NaOH = 0.070 moles HX

Replacing in H-H equation:

pH = 3.14 + log [0.065mol] / [0.070mol]

pH = 3.11 is the pH of the buffer

Plz help Will mark brainliest

Answers

Answer:

H

Explanation:

UMMM H2O .. im assuming the H missing

Concentrated aqueous perchloric acid is 70.5 wt% HClO4 and has a concentration of 11.7 M. Calculate the volume of concentrated perchloric acid that should be diluted to 1.90 L to form a 5.00 M HClO4 solution.

Answers

Answer:

0.812 L

Explanation:

As this is a dilution process problem, we can solve it by using the C₁V₁ = C₂V₂ formula, where in this case:

C₁ = 11.7 MV₁ = ?C₂ = 5.00 MV₂ = 1.90 L

We input the data:

11.7 M * V₁ = 5.00 M * 1.90 L

And solve for V₁:

V₁ = 0.812 L

This means that 0.812 L of concentrated perchloric acid should be diluted to a final volume of 1.90 L.

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