Which of the following is an example of a mixture?

A.

calcium

B.

the atmosphere

C.

carbon dioxide

D.

water

Answer:

A photon

Explanation:

When electrons pass close to the standing wave of light, they hit the particles that are called photons

K20

K = 1+

0 = 2-

K/0 = 1/2

the answer is K:O = 1:2

Answer:

my gues is red not a 100% but its in the 700s

Explanation:

Answer:

[tex]M_{acid}=0.08892M[/tex]

Explanation:

Hello!

In this case, since the reaction between arsenic acid and potassium hydroxide is:

[tex]H_3AsO_4+3KOH\rightarrow K_3AsO_4+3H_2O[/tex]

Thus, since the mole ratio between the acid and the base is 1:3, at the equivalence point we can write:

[tex]3n_{acid}=n_{base}[/tex]

That in terms of molarities and volumes is:

[tex]3M_{acid}V_{acid}=M_{base}V_{base}[/tex]

Thus, the concentration of the arsenic acid is:

[tex]M_{acid}=\frac{M_{base}V_{base}}{3V_{acid}}=\frac{0.1894M*35.21mL}{3*25.00mL} \\\\M_{acid}=0.08892M[/tex]

Best regards!

The concentration (M) of arsenic acid in a solution of 25.00 mL is 0.088M.

Molarity of any solution will be calculated as:

M = n/V, where

V = volume

n is the moles and it will be calculated as:
n = W/M, where

W = given mass

M = molar mass

Given chemical reaction is:

H₃AsO₄ + 3KOH → K₃AsO₄ + 3H₂O

Moles of 35.21 mL of 0.1894 M KOH will be calculated as:

n = (0.1894)(0.035) = 0.0066 moles

From the stoichiometry of the reaction it is clear that:

1 mole of H₃AsO₄ = react with 3 moles of KOH

0.0066 moles of KOH = react with 1/3×0.0066= 0.0022 moles of H₃AsO₄

Now we calculate the molarity of H₃AsO₄ in 25mL of solution as:

M = 0.0022/0.025 = 0.088 M

Hence, required molarity is 0.088 M.

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Answer:

[tex]ionization=1.74\%[/tex]

Explanation:

Hello!

In this case, since the degree of ionization of an acid is computed in terms of the concentration of hydrogen ions and the initial concentration of the acid:

[tex]ionization=\frac{[H^+]}{[HA]} *100\%[/tex]

Because the ionization reaction is represented by:

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

Therefore, the degree or percent ionization turns out:

[tex]ionization=\frac{0.00417M}{0.240M} *100\%\\\\ionization=1.74\%[/tex]

Best regards!

Answer:

Deuterium and Lithium.

Explanation:

I searched it up on google.

Answer:

[tex]V = 0.117 \ L[/tex]

General Formulas and Concepts:

Chem

Explanation:

Step 1: Define

1.57 M KOH (potassium hydroxide)

10.3 g KOH

Step 2: Define conversions

Molar Mass of K - 39.10 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of KOH - 39.10 + 16.00 + 1.01 = 56.11 g/mol

Step 3: Convert

[tex]10.3 \ g \ KOH(\frac{1 \ mol \ KOH}{56.11 \ g \ KOH} )[/tex] = 0.183568 mol KOH

Step 4: Solve for Volume

Step 5: Check

We are given 3 sig figs. Follow sig fig rules.

[tex]0.116922 \ L \approx 0.117 \ L[/tex]

Answer:

The temperature of the water decreases and so the reaction is endothermic

Explanation:

This option is correct because when the 5.0g potassium chloride (KCl) was added to the 100.0ml distilled water, the KCl made the water absorb heat coming from it which made the water decrease in temperature thereby causing an endothermic reaction.

An endothermic reaction actually takes place when the temperature of a system which is isolated decreases. The surroundings of a non-isolated system actually gains heat. While in an exothermic reaction, the temperature of a system increases due to the release or evolution of heat. This heat is released into the surroundings.

Answer:

Kb = 1.6 × 10⁻⁵

Explanation:

Step 1: Given data

Acid dissociation constant of hydrocyanic acid (Ka): 6.2 × 10⁻¹⁰

Concentration of cyanide ion (Cb): 0.1 M

Step 2: Calculate the basic dissociation constant (Kb) of cyanide ion

We have the Ka of HCN. We can calculate the Kb of its conjugate base using the following expression.

Ka × Kb = Kw = 1.0 × 10⁻¹⁴

Kb = 1.0 × 10⁻¹⁴/Ka

Kb = 1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰

Kb = 1.6 × 10⁻⁵

The  Kb for a 0.1 M solution of cyanide ion is :

- 1.6 × 10⁻⁵

Given:

Acid dissociation constant of hydrocyanic acid (Ka)= 6.2 × 10⁻¹⁰

Concentration of cyanide ion (Cb)= 0.1 M

Base dissociation constant (kb)=?

Ka × Kb = Kw = 1.0 × 10⁻¹⁴

Kb = 1.0 × 10⁻¹⁴/Ka

Kb = 1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰

Kb = 1.6 × 10⁻⁵

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Answer: 1M

Explanation:

Molarity = mols/L

moles of lead bromide: 183.51/ 367.0 = 0.5 mol

500 ml/ 1000 mL = .5L

.5 mol / .5 L = 1 mol/L = 1M

Answer:

1.5 moles H₂O

General Formulas and Concepts:

Chemistry - Stoichiometry

Explanation:

Step 1: Define

RxN:   2H₂ + O₂ → 2H₂O

Given:   0.750 moles O₂

Step 2: Stoich

[tex]0.750 \ mol \ O_2(\frac{2 \ mol \ H_2O}{1 \ mol \ O_2} )[/tex] = 1.5 moles H₂O

Step 3: Check

We are given 3 sig figs.

Since our answer is 2 sig figs, we don't need to round.

Answer:

The atmosphere

Explanation:

It reflects and traps the light outside of the Earth's Atmosphere?

I think, hope i helped.

Answer:

Answer. Answer: Nitrogen (N), antimony (Sb), bismuth (Bi) is the order from least conductive to most conductive.

Answer:

nitrogen (N), antimony (Sb), bismuth (Bi)

Answer:

1,002.936 g

Explanation:

The combustion equation of each will be;

CH4 + 2O2 = CO2 + H2O

C3H8 + 5O2 = 3CO2 + 4H2O

We are told the mixture is 344g.

Thus;

For CH4 combustion;

Amount of CH4 = 0.336 × 344 = 115.584g

Molar mass of CH4 is 16 g/mol

Number of moles of CH4 is;

n = 115.584g/(16 g/mol)

n = 7.224 moles

n(CO2) = n(CH4) = 7.224 moles

For C3H8 combustion;

Amount of C3H8 = 0.664 × 344 = 228.416 g

Molar mass of C3H8 = 44 g/mol

Thus;

Number of moles of C3H8 = 228.416 g/(44 g/mol) = 5.19 mol

n(CO2) = 3n(C3H8) = 3 × 5.19 = 15.57 moles

Total moles of CO2 = 7.224 moles + 15.57 moles = 22.794 mol

Molar mass of CO2 = 44 g/mol

Thus amount of CO2 by mass = 22.794 mol × 44 g/mol = 1,002.936 g

Answer:

5 valence elctrons

Explanation:

the outer most orbitals, 3s2 and 3p3 contains 5 electrons, thus valences electrons for P is 5

Answer:

V₂ = 4.4 L

Explanation:

Given data:

Initial volume = 5.5 L

Initial pressure = 650 mmHg (650/760 =0.86 atm)

Initial temperature = 20 °C (20 +273 = 293 K)

Final temperature = 273 K

Final volume = ?

Final pressure =  1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

V₂ =  P₁V₁T₂/T₁ P₂

V₂ = 0.86 atm× 5.5 L × 273 K / 293 k × 1 atm

V₂ = 1291.29 atm.K.L / 293 k.atm

V₂ = 4.4 L

Answer:

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Answer:

0.0148 moles

Explanation:

Solid copper is oxidized with HNO₃:

Cu + 4HNO₃ → Cu²⁺ + 2 NO₃⁻ + 2H₂O + 2NO₂

Where 1 mole of Cu produce 1 mole of Cu²⁺ when nitric acid is in excess.

Moles of Cu that react (Molar mass Cu = 63.546g/mol):

0.94g * (1mol / 63.546g) = 0.0148 moles of Cu

And moles of Cu²⁺ produced are also:

Answer:

Kinetic energy

Explanation:

Fireworks convert chemical energy into kinetic energy to send them flying upward. Hope it is correct!

Answer:

1.44mole of CO

Explanation:

The reaction equation is given as:

         5C  + 2SO₂ →  CS₂ + 4CO

We check to see if the expression is balanced and it is so;

 Now;

 Given;

           1.8mole of C reacted; how many moles of CO are produced;

  From the balanced reaction equation:

             5 mole of C is expected to produce 4 mole of CO

             1.8 mole of C will then produce [tex]\frac{4 x 1.8}{5}[/tex] = 1.44mole of CO

Answer:

The classical cell theory was proposed by Theodor Schwann in 1839. There are three parts to this theory.

Explanation:

Answer:

8.52 m/s

Explanation:

Step 1: Given data

Step 2: Convert "T" to the Kelvin scale

When working with gases, we need to consider the absolute temperature. We will convert from Celsius to Kelvin using the following expression.

K = °C + 273.15 = -16 + 273.15 = 257 K

Step 3: Calculate the root mean square speed of the gas

The root mean square speed measures the average speed of particles in a gas. We will calculate it using the following expression.

[tex]v_{rms} = \sqrt{\frac{3 \times R \times T}{M} } = \sqrt{\frac{3 \times 8.314 J/mol.K \times 257 K}{88.30 g/mol} } = 8.52 m/s[/tex]

Answer:

Butanoic acid.

Explanation:

Answer:

-3

Explanation:

Phosphorus becomes phosphide with a -3 charge

Answer:

0.0099M = [F⁻]

Explanation:

For BaF2, Ksp = 1.7x10⁻⁶

When BaF₂ is in solution, the equilibrium between the solid and the dissociated ions occurs as follows:

BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

Where Ksp = 1.7x10⁻⁶ is defined as:

1.7x10⁻⁶ = [Ba²⁺] [F⁻]²

Where [] are equilibrium concentrations of each ion in solution.

That means you will add F⁻ until its concentration exceeds:

1.7x10⁻⁶ = [0.0173] [F⁻]²

9.827x10⁻⁵ = [F⁻]²

When more F⁻ is added, BaF₂ begins its precipitation.

Answer:

Metalloids and ions

Explanation:

Mercury is not solid at room temperature. (Metals)

Many nonmetals aren't good at conducting electricity. Oxygen for example. (Nonmetals)

Metalloids are always solid at room temp and are fair conductors. (Metalloids)

Ions are same as metalloids. (Ions)

PLS GIVE BRAINLIEST

Answer:

Younger/newer

Explanation:

Stratagraphic means an intrusion or fault is younger than the layer it affects.

Answer:

The correct answer is 4.84

Explanation:

We use the Henderson Hasselbach's equation to calculate the pH of a buffer solution:

pH = pKa = log ( [conjugate base]/[weak acid])

In this case:

conjugate base: acetate ⇒ [conjugate base] =  [acetate] = 0.0200 M

weak acid: acetic acid ⇒ [weak acid] =  [acetic acid] = 0,0200 M

pH = pKa + log ([acetate]/[acetic acid])= pKa + log (0.0200 M/0.0200 M)

When a strong base- such as NaOH- is added, the acid reacts with OH⁻ to form the conjugate base. So, the conjugate base is increased while the acid is decreased. Thus, in 1 liter of solution we have:

acetate = (0,0200 mol/L x 1 L) + (2.0 mmol x 1 mol/1000 mmol) = 0.022 mol

acetic acid= (0,0200 mol/L x 1 L) - (2.0 mmol x 1 mol/1000 mmol) = 0.018 mol

Finally, we calculate the pH:

pH = 4.75 + log (0.022 mol /0.018 mol )= 4.84

Buffer has been the solution that resists the change in the pH of the solution with the addition of acid or base. The pH of buffer with NaOH has been 4.84.

The pH has been the hydrogen ion concentration in the solution. The pH has been equivalent to pKa and can be given as:

[tex]\rm pka=log\dfrac{[base]}{[acid]}[/tex]

The base in the buffer has been the acetate, while acetic acid act as the acid.

The concentration of NaOH added to the solution has been given as:

[tex]\rm Molarity=\dfrac{moles}{volume}\\\\ NaOH=\dfrac{0.002\;mol}{1\;L}\\\\ NaOH=0.002\;M[/tex]

The addition of NaOH adds the concentration of base, and neutralization reduces the concentration of acid. Thus, the new pKa can be given as:

[tex]\rm pKa=log\dfrac{0.2+0.002}{0.2-0.002} \\pKa=log\dfrac{0.022}{0.018}[/tex]

The new pH of the solution has been the sum of the old and the new pKa value.

Thus, the pH of the new buffer has been given as:

[tex]\rm pH=4.75+log\dfrac{0.022}{0.018}\\ pH=4.84[/tex]

The pH of buffer with NaOH has been 4.84.

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