Determine whether the statement about identifying a halide is true: Regardless of any concentration of ammonium solution, the precipitates in the reaction solution of my unknown halide after 0.1M AgNO3 remain because my unknown halide solution contains Br. Select one: True False

Answers

Answer 1

Answer:

False

Explanation:

The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE

This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺


Related Questions

If you drip an ink drop into a cup of water and wait for a few seconds, all the water will be colored with the ink. This experiment is an example of facilitated diffusion ?

true

false ​

Answers

Answer:

false, it is not an example of facilitated diffusion

Answer:

True

Explanation:

When a drop of ink added into the water gradually moves in the whole quantity of water due to this entire water turns into blue color. This is nothing but the diffusion of ink particles into the water molecules. This is because water, as well as ink molecules, are in random motion due to the motion of ink substance.

why is repetition important in science?

Answers

Answer:

the repetition principle is important in scientific research, because the observational indexes are random variables,which requires a certain amount of samples to reveal their changing regularity

repeating something as in a meaning of a word helps us not only remember but have it stuck in our brain. My science teacher in 10th made us rewrite our vocab everyday until all of us knew it. When it came to test taking it was so much easier when I thought of the word. Repetition is so important in science to not only get you to understand the meaning of something but help you remember it.

Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 55.8 g of hydrobromic acid is mixed with 17. g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.

Answers

Answer:

21.4g of HBr is the minimum mass that could be left over.

Explanation:

Based on the reaction:

HBr + NaOH → NaBr + H2O

1 mole of HBr reacts per mole of NaOH

To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:

Moles NaOH -40.0g/mol-

17g * (1mol/40.0g) = 0.425 moles NaOH

Moles HBr -Molar mass: 80.91g/mol-

55.8g * (1mol/80.91g) = 0.690 moles HBr

The difference in moles is:

0.690 moles - 0.425 moles =

0.265 moles of HBr could be left over

The mass is:

0.265 moles * (80.91g/mol) =

21.4g of HBr is the minimum mass that could be left over.

Consider the preparation of methyl benzoate by reacting benzoic acid with methanol using sulfuric acid as a catalyst. Reaction scheme of benzoic acid with methanol, conc. sulfuric acid, and heat over the arrow, and methyl benzoate and water as products. Calculate the molar masses of the reactant and product. Report molar masses to 1 decimal place. Molar mass of benzoic acid g/mol Molar mass of methyl benzoate

Answers

Answer:

See explanation

Explanation:

The molecular mass is the sum of the relative atomic masses of all the atoms in the molecule.

The relative atomic mass of reactants and products are calculated as follows;

Benzoic acid is C7H6O2 hence the molar mass of benzoic acid is ;

7(12) + 6(1) + 2(16) = 84 + 6 + 32 = 122.0 g/mol

Methyl benzoate is C8H8O2

8(12) + 8(1) + 2(16) = 96 + 8 + 32 = 136.0 g/mol

Two methods by which we can conserve water and water the plants.

Answers

Answer:

Two methods which help us to conserve water are:

Sprinkler irrigation system: this irrigation has an arrangement of vertical pipes with rotating nozzles on the top. It is more useful in the uneven and sandy land where sufficient water is not available.

Drip irrigation system: this irrigation system has an arrangement of pipes or tubes with very small holes in them to water plants drop by drop just at the base of the root. It is very efficient as water is not wasted at all.

When a marble is dropped into a beamer of water

Answers

Answer:

The water will rise.

Explanation:

hope this helps you

-Sweety<3

The mass of the marble is greater than that of the water. The marble weighs more than an equivalent volume of the water. The force from dropping the marble breaks the surface tension of the water. The marble has greater mass and volume than the water.

Is Water and kerosine a mixture

Answers

Answer:

No.Kerosene oil and water do not mix with each other and form two separate layers.

Answer:

No

Explanation:

They cannot be mixed together they will form upper and lower layer

A 18.0 L gas cylinder is filled with 6.20 moles of gas. The tank is stored at 33 ∘C . What is the pressure in the tank?
Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

8.65 atm

Explanation:

Using ideal law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (Latm/molK)

T = temperature (K)

According to the information given in this question;

V = 18.0 L

n = 6.20 moles

R = 0.0821 Latm/molK

T = 33°C = 33 + 273 = 306K

P = ?

Using PV = nRT

P × 18 = 6.20 × 0.0821 × 306

18P = 155.76

P = 155.76/18

P = 8.65 atm

who much the velocity of a body when it travels 600m in 5 min​

Answers

Answer:

2 m/s

Explanation:

Applying the formulae of velocity,

V = d/t............. Equation 1

Where V = Velocity of the body, d = distance, t = time

From the question,

Given: d = 600 m, t = 5 minutes = (5×60) = 300 seconds.

Substitute these values into equation 1

V = 600/300

V = 2 m/s.

Hence the velocity of the body when it travels is 2 m/s

what is a compound?And what are some common examples of a compound?​

Answers

Answer:

Compounds are substances made from atoms of different elements joined by chemical bonds. Common examples are water (H2O), salt (sodium chloride, NaCl), methane (CH4).

An ionic bond is a bond

Answers

Answer:

That involve the complete transfer of an electron from one atom of an element to another

A buffer is a solution that: Select one: a. Results from mixing a strong acid and a strong base. b. When added to another solution, decreases the pH. c. When added to another solution, increases the pH. d. Prevents a drastic change in pH when an acid or base enters a solution.

Answers

Answer:

d. Prevents a drastic change in pH when an acid or base enters a solution.

Explanation:

The purpose of a buffer is to resist pH change and keep the solution relatively stable.

Analyze the data and determine the actual concentration of calcium chloride in the solution. Show all calculations and report in % wt/v concentration.
Known; Mass of CaCl2 present in original solution, based on actual yield= 1.77g moles
CaCl2 present in original solution, based on actual yield= 1.77g/molar mass of CaCl2=1.77g/110.98g/mol=0.016 moles
Total Volume of solution =V, which is 80ml

Answers

Answer:

2.21% wt/v

Explanation:

The mass/volume percentage, %wt/v, is an unit of concentration used in chemistry defined as 100 times the ratio of the mass of solute in g (In this case, CaCl2 = 1.77g) and the volume of solution in mL = 80mL

The %wt/v of this solution is:

%wt /v = 1.77g / 80mL * 100

%wt/v = 2.21% wt/v

9.7300x10^2 + 9.8700x10^3

? × 10^?

Answers

Answer:

19.6 x 10⁵

1.96 x 10⁶

I hope it's helps you

A chemist is preparing to carry out a reaction that requires 5.75 moles of hydrogen gas. The chemist pumps the hydrogen into a 10.5 L rigid steel container at 20.0 °C. To what pressure, in kPa, must the hydrogen be compressed? (Show all work for full credit and circle your final answer) *

Answers

Answer:

The hydrogen must be compressed to 1333.13302 kPa.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

P= ?V= 10.5 Ln= 5.75 molesR= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 20 C= 293 K (being 0 C= 273 K)

Replacing:

P* 10.5 L= 5.75 moles* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] * 293 K

Solving:

[tex]P=\frac{5.75 moles* 0.082 \frac{atm*L}{mol*K} * 293 K}{10.5 L}[/tex]

P= 13.157 atm

If 1 atm is equal to 101.325 kPa, then 13.157 atm is equal to 1333.13302 kPa.

The hydrogen must be compressed to 1333.13302 kPa.

According to the EPA Lead and Copper Rule (LCR), the action level for Pb in drinking water (the level at which threat to human health requires public notification and action towards mitigation) is 15 ppb. If you were to add enough phosphate to the system
saturated with respect to Pb3(PO4)2(s), would the [Pb2+] be below the action limit?

Answers

Answer:

The right answer is "105.17 ppb".

Explanation:

According to the question,

The amount of [tex]Pb^{2+}[/tex] in ppb will be:

= [tex]0.5076\times 10^{-6}\times 207.2\times 106[/tex]

= [tex]105.17 \ ppb[/tex]

Thus, the amount of [tex]Pb^{2+}[/tex] is above action limit.

Assume that all products containing “Bromide” or an ingredient ending in “-ol” are toxic.

Answers

Product B and E are toxic because they contain "BROMIDE" or ingredients that end in 'ol'

Based on the directions given in the information of this question, any product containing "bromide" or containing an ingredient that ends in "ol" is assumed to be TOXIC.

After carefully evaluating the ingredient contents of each product in the image attached to this question, it was realized that product B contains "pyridostigmine bromide" as an ingredient while product E contains "butorphanol" as an ingredient. Hence, in accordance to the guide given in this question, products B and E are toxic.

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The mole fraction of NaCl in an
aqueous solution is 0.132. How
many moles of NaCl are present in
1 mole of this solution?
Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol

Answers

Answer:

Moles of water are 0.868

Explanation:

A cyclopropane-oxygen mixture is used as an anesthetic. If the partial pressure of cyclopropane in the mixture is 330 mmHg and the partial pressure of the oxygen is 1.0 atm, what is the total pressure of the mixture in torr

Answers

Answer:

1090 Torr

Explanation:

Step 1: Given data

Partial pressure of cyclopropane (pC₃H₆): 330 mmHg (330 Torr)Partial pressure of oxygen (pO₂): 1.0 atm

Step 2: Convert pO₂ to Torr

we will use the conversion factor 1 atm = 760 Torr.

1.0 atm × 760 Torr/1 atm = 760 Torr

Step 3: Calculate the total pressure of the mixture (P)

The total pressure of the mixture is the sum of the partial pressures of the gases.

P = 330 Torr + 760 Torr = 1090 Torr

Hydrogen chloride decomposes to form hydrogen and chlorine, like this:

2HCl(g) + H2(g) â Cl2(g)

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen chloride, hydrogen, and chlorine has the following composition:

compound pressure at equilibrium
HCl 84.4 atm
H2 77.9 atm
Cl2 54.4

Required:
Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answers

Solution :

Given :

Partial pressure of HCl, [tex]$P_{HCl}$[/tex] = 84.4 atm

Partial pressure of [tex]H_2[/tex], [tex]$P_{H_2}$[/tex] = 77.9 atm

Partial pressure of [tex]Cl_2[/tex], [tex]$P_{Cl_2}$[/tex] = 54.4 atm

Reaction :

[tex]$2HCl (g) \leftrightharpoons H_2(g) + Cl_2(g)$[/tex]

Using equilibrium concept,

[tex]$k_p=\frac{(P_{H_2})(P_{Cl_{2}})}{(P_{HCl})^2}$[/tex]

[tex]$k_p=\frac{77.9 \times 54.4}{(84.4)^2}$[/tex]

[tex]$k_p=0.594$[/tex]

[tex]k_p=0.59[/tex]  (in 2 significant figures)

or [tex]k_p=5.9 \times 10^{-1}[/tex]

A fusion reaction releases energy because the binding energy of the resulting nucleus:______.
a. is released in the process.
b. is equal to the binding energy of the original nuclei.
c. is absorbed in the process.
d. is less than the binding energy of the original nuclei.
e. is greater than the binding energy of the original nuclei.

Answers

Answer:

a. is released in the process

Explanation:

In fusion reaction the nucleus is unstable so it releases its binding energy resulting in decreasing its mass so it becomes more stable.

Write the chemical formula for the following:

a. The conjugate acid of amide ion, NH₂-
b. The conjugate base of nitric acid, HNO₃

Answers

Follow the rules of Bronsted Lowry theory

Acids take a HBases donate a H

So

#a

NH_2-

Add a H

Conjugate acid is NH_3

#b

HnO_3

Take a H

Conjugate base is NO_3-

#1

Conjugate acid means one H+ is added

NH_2+H+=NH_3

sw

#2

Conjugate base means donate a H+

HNO_3-H=NO_3-

Draw a relative energy diagram showing a conformational analysis of 2,2-dichloropropane along C1-C2 bond. Clearly label all staggered conformations and all eclipsed conformations with the corresponding Newman projections.

Answers

Answer:

^#))(899900000)My020

how many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3 ?​

Answers

Answer:

Explanation:

The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.

0.2175,0.0725 moles of [tex]H_2[/tex] and [tex]N_2[/tex] can be formed by the decomposition of 0.145 mol of ammonia, [tex]NH_3[/tex]

The reaction for the decomposition of ammonia is as follows:-

[tex]N_2+3H_2 \rightarrow 2NH_3[/tex]

Calculate the mole  of [tex]H_2[/tex] and [tex]N_2[/tex] as follows:-

[tex]Mole\ of \ H_2=0.145\ mol\ NH_3\times\frac{3\ mol\ H_2}{2\ mol\ NH_3} \\\\=0.2175\ mol\ H_2[/tex]

[tex]Mole\ of \ N_2=0.145\ mol\ NH_3\times\frac{1\ mol\ N_2}{2\ mol\ NH_3} \\\\=0.0725\ mol\ H_2[/tex]

Hence, the number of moles of [tex]H_2[/tex] and [tex]N_2[/tex]  are 0.2175 mol, and 0.0725 mol.

To know more about:-

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Leo carefully pipets 50.0 mL of 0.500 M NaOH into a test tube. She places the test tube
into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.250 M HCl into
another test tube. When Leo reaches to put this test tube of acid into the beaker along
with test tube of base she accidentally knocks the test tubes together hard enough to
break them and their respective contents combine in the bottom of the beaker. Is the
solution formed from the contents of the two test tubes acidic or basic? What is the pH of
the resulting solution?
Please answer below questions one by one to assist you receive full credits
(Alternatively, you can discard my hints below, solve the problem using your own way
and send me the picture/copy of your complete work through email)
The mole of NaOH before mixing is
mol (save 3 significant figures)
The mole of HCl before mixing is
mol (save 4 significant figures)
After mixing, the solution is
(choose from acidic or basic)
The total volume of mixture is
L (save 3 significant figures)
The concentration of [OH-] is
M (save 3 significant figures)
The concentration of [H'l is
M (save 3 significant figures)

Answers

Let's consider the neutralization reaction between HCl and NaOH.

NaOH + HCl ⇒ NaCl + H₂O

To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.

NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol

HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol

Now, let's determine the reactant in excess and the remaining moles of that reactant.

                    NaOH    +     HCl ⇒ NaCl + H₂O

Initial           0.0200       0.0188

Reaction    -0.0188       -0.0188

Final         1.20 × 10⁻³          0

The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³  moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:

[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M

NaOH is a strong base according to the following equation.

NaOH ⇒ Na⁺ + OH⁻

The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.

The pOH is:

pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02

We will calculate the pH using the following expression.

pH = 14.00 - pOH = 14.00 - 2.02 = 11.98

The pH is 11.98. Since pH > 7, the solution is basic.

You can learn more about neutralization here: https://brainly.com/question/16255996

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates. The chemical equation for this reaction is

Answers

Answer:

p'PCl3 =  6.8 torr

p'Cl2 =26.4 torr

p'PCl5 =223.4 torr

Explanation:

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates. The chemical equation for this reaction is

PCl3(g) + Cl2(g) ---> PCl5(g)

Calculate the new partial pressures after equilibrium is reestablished. [in torr]

pPCl3

pCl2

pPCl5

Step 1: Data given

Partial pressure before adding chlorine gas:

Partial pressure of PCl5 = 217.0 torr

Partial pressureof PCl3 = 13.2 torr

Partial pressureof Cl2 = 13.2 torr

A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing

Step 2: The equation

PCl3(g)+Cl2(g) ⇔ PCl5(g)

Step 3: The expression of an equilibrium constant before adding chlorine gas

Kp = pPCl5 / (pPCl3 * pCl2)

Kp = 217.0 / (13.2 * 13.2)

Kp =  1.245

Step 4:  The expression of an equilibrium constant after adding chlorine gas

Partial pressure of PCl5 = 217.0 torr

Partial pressure of PCl3 = 13.2

Partial pressure of Cl2 = TO BE DETERMINED

Step 5: The total pressure of the system

Ptotal = pPCl5 + pPCl3 + pCl2

263.0 torr = 217.0 torr + 13.2 torr + pCl2

pCl2 = 263.0 - 217.0 -13.2 = 32.8 torr

Step 6: The initial pressure

The equation: PCl3(g)+Cl2(g) ⇔ PCl5(g)

pPCl3 = 13.2 torr

pCl2 = 32.8 torr

pPCl5 = 217.0 torr

Step 7: The pressure at the equilibrium

p'PCl3 = (13.2 -x) torr

p'Cl2 = (32.8 - x) torr

p'PCl5 = (217.0 + x) torr

Step 8: The equilibrium constant

'Kp =  p'PCl5 / (p'PCl3 * p'Cl2)

1.245 = (217.0+x) / ((13.2-x)(32.8-x)

x = 6.40 torr

p'PCl3 = 13.2 -6.40 = 6.8 torr

p'Cl2 = 32.8 - 6.40 =26.4 torr

p'PCl5 = 217.0 + x) 6.4 = 223.4 torr

Calculate the volume of a 89.51 g sample of carbon dioxide at 281.8 K and 843.9 torr. Round your answer
to the nearest L. Do not include units.

Answers

Answer:

1,000.000

Explanation:

Ortho and para hydrogen are....... a). molecular form. b). Nuclear form. c) allotropic form. d). All​

Answers

Ortho and para hydrogen are nuclei forms

write anode and cathode in Zn-Ag galvanic cell​

Answers

Explanation:

Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).

By convention in standard cell notation, the anode is written on the left and the cathode is written on the right. So, in this cell: Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).

Consider the following titration for these three questions:

1.00 L of 2.00 M HCl is titrated with 2.00 M NaOH.

a. How many moles of acid are equal to one equivalent in this titration?
b. How many moles of HCl are found in solution at the halfway point of the titration?
c. How many liters of base will be needed to reach the equivalence point of the titration?

Answers

Answer:

a. 1 mole of acid is equal to one equivalent.

b. 1.00 moles of HCl are found.

c. 1L of 2.00M NaOH is needed to reach the equivalence point

Explanation:

HCl reacts with NaOH as follows:

HCl + NaOH → NaCl + H2O

Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1

a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent

b. The initial moles of HCl are:

1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl

At the halfway point, the moles of HCl are the half, that is:

1.00 moles of HCl are found

c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:

2.00moles NaOH * (1L / 2.00mol) =

1L of 2.00M NaOH is needed to reach the equivalence point

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