A type of matter formed by chemically joining two or more elements is known as a A compound B) solution c) colloid D suspension
​

Answer:

2.56g of H₂S

Explanation:

The reaction of H₂S with Pb(CH₃COO)₂ is:

H₂S + Pb(CH₃COO)₂ → 2 CH₃COOH + PbS

Where 1 mole of H₂S produce 1 mole of PbS

18.00g of PbS (Molar mass: 239.3g/mol):

18.00g of PbS ₓ (1mol / 239.3g) = 0.0752 moles of PbS

As 1 mole of PbS is produced from 1 mole of H₂S, moles needed of H₂S are 0.0752

As molar mass of H₂S is 34.1g/mol, mass in 0.0752moles is:

0.0752 mol H₂S ₓ (34.1g / mol) = 2.56g of H₂S

One liquid substance that could be used in the laboratory for dissolving dry mortar on floor tiles is hydrochloric acid (HCl).

Hydrochloric acid is a clear, colorless, highly corrosive solution, usually sold in concentrations ranging from 28% to 35%. Hydrochloric acid can be used to dissolve or break down the mortar on the floor tiles, it reacts with the calcium carbonate in the mortar and dissolves it, making it easy to wipe away. It is also helpful in removing rust and hard water stains from concrete and ceramic surfaces.

However, hydrochloric acid is extremely corrosive, so safety precautions should be taken when handling it. Proper personal protective equipment, such as gloves and eye protection, should be worn, and it should be used in a well-ventilated area to prevent inhalation of harmful fumes. So therefore HCl or hydrochloric acid is one liquid substance that could be used in the laboratory for dissolving dry mortar on floor tiles.

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Answer:

The new moles of the gas in the flask is 2.13 moles.

Explanation:

Given;

number of moles of gas, n = 1.9 mol

temperature of the gas, T = 21 °C  = 21 + 273 = 294 K

pressure of gas, P = 697 mmHg

volume of gas, V = ?

Apply ideal gas law;

PV = nRT

Where;

R is gas constant, = 62.363 mmHg.L / mol. K

V = nRT / P

V = (1.9 x 62.363 x 294) / 697

V = 49.98 L

New pressure of the gas, P = 795 mmHg

New temperature of the gas, T = 26 °C = 273 + 26 = 299 K

New moles of the gas, n = ?

Volume of the gas is constant because volume of the flask is the same when more gas was added.

n = PV / RT

n = (795 x 49.98) / (62.363 x 299)

n = 2.13 moles

Therefore, the new moles of the gas in the flask is 2.13 moles.

Answer:

Calcium hope it helped

Explanation:

Answer:

i believe its gas sorry if i am wrong

Explanation:

Answer:

Geothermal energy is an uptrending alternative source of production that relies on the earth's crust and elements of radioactivity.

Answer:

It is carried to the lungs to be exhaled

Explanation:

Answer:

I dont know

Explanation:

I dont know

Answer: \(1.6\times 10^4\)

Explanation:

Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example : 5000 is written as \(5.0\times 10^3\)

In division , the result would contain the same number of significant digits as there in the least precise number , thus the answer would have 2 significant digits.

\(\frac{1.248\times 10^{-2}}{7.8\times 10^{-7}}=1.6\times 10^4\)

When helium–neon laser light is sent through a 0.310-mm-wide single slit, the width of the central maximum on a screen 3.00 m from the slit is approximately 12.26 mm.

In the case of a single-slit diffraction pattern, the central maximum is the most prominent maximum that lies directly in front of the slit, with a width of its own. It is the brightest and most prominent feature in the pattern as it is the only one with a maximum intensity. It is also known as the principal maximum or the zeroth order maximum.

To find the width of the central maximum, we can use the formula for single-slit diffraction:

Width = 2 * L * λ / a

where L is the distance from the slit to the screen (3.00 m), λ is the wavelength of the helium-neon laser light (632.8 nm = 632.8 x 10⁻⁹ m), and a is the width of the slit (0.310 mm = 0.310 x 10⁻³ m).

Width = 2 * 3.00 * (632.8 x 10⁻⁹) / (0.310 x 10⁻³)

Width = 6 * (632.8 x 10⁻⁹) / (0.310 x 10⁻³)

Width ≈ 0.01226 m, or 12.26 mm

So, the width of the central maximum on the screen is approximately 12.26 mm.

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The expected ground-state electron configuration for the element with one unpaired 5p electron that forms a covalent compound with fluorine is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^2.

To determine the expected ground-state electron configuration, we need to consider the element with one unpaired 5p electron that forms a covalent compound with fluorine.

Since we are looking for an element with one unpaired 5p electron, we can refer to the periodic table.

The element in question is in the p-block of the periodic table, and its electron configuration can be written as:

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^1

To determine the expected ground-state electron configuration, we need to understand that elements tend to achieve a stable configuration by filling their valence shell. In this case, the valence shell is the 5p orbital.

To form a covalent compound with fluorine, the element must gain one electron from fluorine.

This electron would occupy one of the unpaired 5p orbitals, resulting in the following electron configuration:

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^2

Therefore, the expected ground-state electron configuration for the element with one unpaired 5p electron that forms a covalent compound with fluorine is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^2.

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We recover 4.583 gm of the compound in the ether when the compound has a distribution coefficient of 11.

Let S be the compound

Vaq = volume of water = 20ml

Vether = volume of ether = 20 ml

Distribution coeffiecient D = S_ether/ S_ water

assuming the compound was in water initially and using conservation of mass

moles of S in water initially = moles of S in water later + moles of S in ether

[Saq]1 = (mol Saq)1/ Vaq               (1)

[Sether] 1 = (mol Sether)1 / Vether     (2)

putting equation 2 in 1

[Sether]1 = (mol Saq)0  - (mol Saq)1 / Vether

(mol Saq)1 / (mol Saq)0 = Vaq / D. Vether + Vaq

which gives the fraction of S that remains in water after 1st extraction

(q_aq)1 = Vaq / D. Vether + Vaq = 20 / 11*20 + 20 = 0.0833

The amount of compound we will recover in ether = (1 - q_aq) * ( amt. of the compound) = (1 - 0.0833) * ( 5) = 4.583 gm

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We should combine 480.2 mL of butanol with 30.0 mL of ethanol to make a 5.88% volume percent solution of ethanol in butanol.

To find the volume of butanol needed to make a 5.88% volume percent solution of ethanol in butanol, follow these steps:

Step 1: Identify the given information.
- Volume of ethanol (V₁) = 30.0 mL
- Percent volume of ethanol in the solution (P) = 5.88%

Step 2: Express the volume percent as a decimal.
P = 5.88% ÷ 100 = 0.0588

Step 3: Write the formula for volume percent.
Volume percent (P) = (Volume of ethanol (V₁)) ÷ (Total volume (V_total))

Step 4: Rearrange the formula to find the total volume (V_total).
V_total = V₁ ÷ P = 30.0 mL ÷ 0.0588 ≈ 510.2 mL

Step 5: Calculate the volume of butanol (V₂).
Volume of butanol (V₂) = Total volume (V_total) - Volume of ethanol (V₁) = 510.2 mL - 30.0 mL = 480.2 mL

So, we should combine 480.2 mL of butanol with 30.0 mL of ethanol to make a 5.88% volume percent solution of ethanol in butanol.

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The statement that would best explain the observation of no liquid being evident in the sealed bottle is: b. Liquid and vapor are at equilibrium in the bottle.

When liquid and vapor are at equilibrium in a closed system, it means that the rate of condensation (liquid turning into vapor) is equal to the rate of vaporization (vapor turning into liquid). In this case, it appears that all the liquid has vaporized, and no liquid is evident. This suggests that the liquid and vapor have reached a state of equilibrium, where the amount of liquid remaining is negligible compared to the amount of vapor present.

The vapor state is favored when equilibrium is established because the pressure exerted by the vapor phase reaches a point where it equals the vapor pressure of the liquid at that temperature. At this equilibrium point, no further net condensation or vaporization occurs, resulting in the absence of visible liquid in the sealed bottle.


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Any hydrocarbon may be completely burned to produce carbon dioxide and water.

Octane, or C8H18, is a component of the fuel that powers your automobile. Octane burns to produce CO2 and water when it is burnt. The availability of oxygen is the primary component in the process. In an oxygen-free environment, combustion is impossible.

2 water molecules to 2 carbon dioxide molecules make up the molar ratio. For every mole of water in the molecule H2O, there are two moles of hydrogen atoms.

Carbon monoxide (CO) and water are byproducts of the incomplete combustion of octane, C8H18.

Carbon dioxide and water are the byproducts of a complete combustion process of any fuel.

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Answer:

32

Explanation:

Answer:

d nitric acid

Explanation:

hope it helps

Explanation:

a. Hydrochloric acid is not a monobasic acid

The empirical formula of a compound is the simplest whole-number ratio of its atoms. Based on the percent composition of the compound, the empirical formula can be calculated as follows:

C: 40% = 4 atoms

H: 6.7% = 0.67 atoms (rounded to 1 atom)

O: 53.3% = 5.33 atoms (rounded to 6 atoms)

The empirical formula is therefore C2H4O2.

The empirical formula of a compound can be calculated from its elemental composition by dividing the number of atoms of each element by the greatest common factor. The resulting formula represents the smallest whole number ratio of atoms in the compound. For example, if a compound is 40% carbon, 6.7% hydrogen, and 53.3% oxygen, the number of atoms can be calculated as follows: 40 g of carbon corresponds to 10 moles, 6.7 g of hydrogen corresponds to 1 mole, and 53.3 g of oxygen corresponds to 6.5 moles. The empirical formula is then C2H4O, which represents the smallest whole number ratio of atoms in the compound.

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(a)

pH = 4.77

; (b)

[

H

3

O

+

]

=

1.00

×

10

-4

l

mol/dm

3

; (c)

[

A

-

]

=

0.16 mol⋅dm

-3

Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

m

m

m

m

m

m

m

m

l

HA

m

+

m

H

2

O

⇌

H

3

O

+

m

+

m

l

A

-

I/mol⋅dm

-3

:

m

m

0.05

m

m

m

m

m

m

m

m

l

0

m

m

m

m

m

l

l

0

C/mol⋅dm

-3

:

m

m

l

-

x

m

m

m

m

m

m

m

m

+

x

m

l

m

m

m

l

+

x

E/mol⋅dm

-3

:

m

0.05 -

l

x

m

m

m

m

m

m

m

l

x

m

m

x

m

m

m

x

K

a

=

[

H

3

O

+

]

[

A

-

]

[

HA

]

=

x

2

0.05 -

l

x

=

3.27

×

10

-4

Check for negligibility

0.05

3.27

×

10

-4

=

153

<

400

∴

x

is not less than 5 % of the initial concentration of

[

HA

]

.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x

2

0.05

−

x

=

3.27

×

10

-4

x

2

=

3.27

×

10

-4

(

0.05

−

x

)

=

1.635

×

10

-5

−

3.27

×

10

-4

x

x

2

+

3.27

×

10

-4

x

−

1.635

×

10

-5

=

0

x

=

1.68

×

10

-5

[

H

3

O

+

]

=

x

l

mol/L

=

1.68

×

10

-5

l

mol/L

pH

=

-log

[

H

3

O

+

]

=

-log

(

1.68

×

10

-5

)

=

4.77

(b)

[

H

3

O

+

]

at pH 4

[

H

3

O

+

]

=

10

-pH

l

mol/L

=

1.00

×

10

-4

l

mol/L

(c) Concentration of

A

-

in the buffer

We can now use the Henderson-Hasselbalch equation to calculate the

[

A

-

]

.

pH

=

p

K

a

+

log

(

[

A

-

]

[

HA

]

)

4.00

=

−

log

(

3.27

×

10

-4

)

+

log

(

[

A

-

]

0.05

)

=

3.49

+

log

(

[

A

-

]

0.05

)

log

(

[

A

-

]

0.05

)

=

4.00 - 3.49

=

0.51

[

A

-

]

0.05

=

10

0.51

=

3.24

[

A

-

]

=

0.05

×

3.24

=

0.16

The concentration of

A

-

in the buffer is 0.16 mol/L.

hope this helps :)

The equilibrium constant of this reaction is 1.80×10-5

Given data,

pH of solution = 2.873

Number of moles of acetic acid (m) = 0.05 moles

Volume of water (V) =  500 mL = 0.5L

So, concentration (C) = m/V in lit = 0.05/0.5 = 0.1 M

Equilibrium constant ( K ) = \([CH_{3} COO-]\)×\([H+_{} ]\)/\([CH_{3} COOH]\)

Since, acetic acid is weak acid,

So, Equilibrium constant ( K ) = \([H+]^{2}\)/\([CH_{3} COOH]\)   ....(i)

As the pH = 2.873, the \([H+_{} ]\) is antilog of -2.873 or 1.34×10-3 M.

Putting the value of concentration of \(H+_{}\) and \(acetic_{} acid\) in equation (i).

Equilibrium constant ( K ) = 1.80×10-5

The acid which is partially dissociates into ions on dissolving in aqueous solution is called weak acid.

Example: \(acetic_{} acid\).

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The temperature of 16-gram sample of oxygen gas with a volume of 11.2 liters at standard pressure is 272.83K.

The volume of an ideal gas can be calculated using the ideal gas law equation:

PV = nRT

Where;

1 × 11.2 = 0.5 × 0.0821 × T

11.2 = 0.04105T

T = 11.2/0.04105

T = 272.83K

Therefore, the temperature of 16-gram sample of oxygen gas with a volume of 11.2 liters at standard pressure is 272.83K.

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Answer:

0.233M

Explanation:

M1 = 0.700M

V1 = 2L

V2 = 3L

M2 = ?

M1V1 = M2V2

(0.700)(2) = M2(3)

M2 = 0.233M

Answer:

B. Impulse

Explanation:

A= she is not resisting buying the car

B= She bought the car right away(impulse)

C = She did not plan ahead on what car she would buy

D= Not reasonable

After addition of HCl in the equation, pH = 9.21 and After addition of NaOH in the solution, pH = 9.31 .

Given the buffer contains 50 mL of 0.3 M NH₃

therefore moles of NH₃ = 50 mL × 0.3 mol / 1000 mL

                                  = 0.0150 moles.

similarly moles of NH₄Cl = 0.015 moles.

The addition of 7.5 mL of 0.125 M HCl = 7.5 mL × 0.125 mol / 1000 mL

                                         = 9.375 × 10⁻⁴ moles of HCl.

The added HCl will react with NH₃ to form NH₄Cl

                            NH₃        +      HCl           1         NH₄Cl

initial                      0.015           0.0009375                  0.015

change                   -0.0009375     - 0.0009375               + 0.0009375

final                       0.0140625               0                         0.0159375

The final volume of the solution = 100 mL + 7.5 mL = 107.5 mL

                                             = 0.1075 L

Therefore the concentration  of NH₃ after addition of HCl =  mol / L

                                    = 0.0140625 mol / 0.1075 L

                                                  = 0.131 M

  similarly the conc of NH₄Cl after addition of HCl = mol / L

                                     = 0.0159375 mol / 0.1075 L

                                               = 0.148 M

By Henderson equation,

pOH = pKb + log [salt/base]

     = 4.74 + log [0.148 / 0.131]

       = 4.74 + 0.053

       = 4.793

pH = 14  - pOH = 14 - 4.793

     = 9.21

The addition of 7.5 mL of 0.125 M NaOH = 7.5 mL  × 0.125 mol / 1000 mL

                                    = 9.375 × 10⁻⁴ moles of NaOH.

The added NaOH will react with NH₄Cl  to form NH₃

                 NH₄Cl        +     NaOH          1    NH₃   + NaCl + H₂O

initial       0.015                0.0009375                      0.015

change                  -0.0009375        - 0.0009375               + 0.0009375

final                       0.0140625               0                         0.0159375

The final volume of the solution = 100 mL + 7.5 mL = 107.5 mL

                                             = 0.1075 L

Therefore the  concentration  of NH₃ after addition of NaOH =  mol / L

                                                    = 0.0159375 mol / 0.1075 L

                                                   = 0.148 M

 similarly the concentration  of NH₄Cl after addition of NaOH = mol / L

                                                      = 0.0140625 mol / 0.1075 L

                                                       = 0.131 M

By Henderson equation,

pOH = pKb + log [salt/base]

      = 4.74 + log [0.131 / 0.148]

       = 4.74 - 0.053

       = 4.69

pH = 14  - pOH = 14 - 4.693

     = 9.31

Thus

After addition of HCl in the equation :

[NH₃] = 0.131

[NH₄Cl] = 0.148

pH = 9.21

After addition of NaOH in the solution :

[NH₃] = 0.148

[NH₄Cl] = 0.131

pH = 9.31

A buffer is a solution that can resist changing its pH when acidic or basic ingredients are added. It can neutralize small amounts of added acid or base, maintaining a relatively stable pH in the solution. This is significant for processes and additionally responses which require explicit and stable pH ranges.

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Answer:

3 and 6 (weight change and nothing changes)

Explanation:

Hi there!

A chemical change is a change when matter changes into a new substance and has a new chemical property.

The signs that a chemical change is taking place are:

1. change in color

2. change in smell

3. change in energy (for example, there is a change in temperature (thermal energy))

4. a gas is formed (fizzing/bubbling/foaming are signs of this!)

5. formation of a solid (a precipitate)

A physical change is any change to the size, shape, or state of a substance. The substance still has the same chemical property.

these are your eight options:

1. tarnished is formed (a sign of a chemical change; a change in color)

2. gas is produced (a sign of a chemical change)

3. weight change (NOT a sign of a chemical change. Even though the weight changed, the substance still has the same chemical composition and therefore, there wasn't any change to its chemical identity)

4. Temperature change (a sign of a chemical change; a change in energy)

5. color change (a sign of a chemical change)

6. nothing changes (NOT a sign of a chemical change. If nothing happens, the substance is still the same as it was originally and there was no change to its chemical identity.)

7. forms a precipitate (a sign of a chemical change)

8. energy is released (a sign of a chemical change)

therefore the two options that are NOT a sign of a chemical change are 3 and 6 (weight change and nothing changes)

Hope this helps! :)

The concentration of the hydronium ion in hydrochloric acid is 0.0045 M, and the pH of the solution is 2.34.

pH is the potential of the hydrogen or the hydronium ions in the aqueous solution.

As the solution contains \(4.5 \times 10^{-3} \;\rm M\;\) HCl the concentration of the hydronium ion will be the same, \(4.5 \times 10^{-3} \;\rm M.\)

The pH of the solution is calculated as:

\(\begin{aligned} \rm pH &= \rm -log[H^{+}]\\\\&= - \rm log (4.5 \times 10^{-3})\\\\&= 2.34\end{aligned}\)

The concentration of the hydroxide ion is calculated from pH and hydronium ion as:

\(\begin{aligned} \rm [H_{3}O^{+}][OH^{-}] &= 10^{-14}\\\\&= \dfrac{1 \times 10^{-14}}{4.5 \times 10^{-3}}\\\\&= 2.2 \times 10^{12}\end{aligned}\)

Now, for the calcium hydroxide solution, the calculations are shown as,

\(\begin{aligned} \rm (H_3}\rm O^{+}) &= \rm antilog (-pH)\\\\&= \rm antilog (-8)\\\\&= 10^{-8} \;\rm M\end{aligned}\)

pOH is calculated as:

\(\begin{aligned} \rm pOH &= 14- 8 = 6\\\\\rm [OH^{-}] &= \rm antilog (-6)\\\\&= 10^{-6} \end{aligned}\)

The concentration of calcium hydroxide is calculated as:

\(\begin{aligned} &= \dfrac{1}{2} \times \rm [OH^{-}]\\\\&= 5 \times 10^{-4} \;\rm M\end{aligned}\)

Therefore, the pH and the pOH give the concentration of the hydrogen or the hydronium ion and the hydroxide ion.

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700.4 K is the e new temperature needed for the gas to increase its volume to 30.5 L at a constant pressure which is determined using  Charle's law.

The initial volume of gas = 10.81 L

Temperature -25°C

The final volume of gas = 30.5 L

Here we can use Charle's law because of changes in the volume and temperature of a gas at constant pressure.

The relation between volume and temperature is written as:

V1/T1 = V2/T2

We need to convert the temperature from the Celsius scale to the Kelvin scale.

T1 = -25°C + 273.15

T1 = 248.15 K

Substituting the values, we get:

V1/T1 = V2/T2

T2 = (V2 * T1) / V1

T2 = (30.5 L * 248.15 K) / 10.81 L

T2 = 700.4 K

Therefore, we can infer that the new temperature needed is 700.4 K.

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The first and second ionization of a diprotic acid cannot be treated as completely separate reactions when the reaction is taking place in an environment with a fixed pH.

The second ionization of the acid is dependent on the concentration of the ions produced from the first ionization.

If the pH is fixed, then the concentration of the first ionization is also fixed, so the second ionization will not occur completely independently.

For example, a diprotic acid such as oxalic acid can be completely ionized in two steps. In the first ionization, the hydrogen ions of the oxalic acid are replaced with hydroxide ions, forming the oxalate ion:

H2C2O4 + 2H2O → H3O+ + HC2O4–

In the second ionization, the oxalate ion is further dissociated, forming two separate anions and hydronium ions:

HC2O4– + H2O → H3O+ + C2O4–2

However, in an environment with a fixed pH, the second ionization will not take place as the concentration of oxalate ions from the first ionization is fixed.

Therefore, the two ionizations must be treated together in order to accurately predict the final concentrations of the products.

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Answer:

The first ionization constant is greater than the second ionization constant by only a factor of 10.

Explanation:

The two ionization constants must differ by a factor of at least 20 in order to treat the first and second ionizations as chemically (and mathematically) distinct.