If there are a 1000 ml per 1 L and a 1000g per kilogram
a. How many ml are there in 5.0 L?
b. How many kg are there in 230g?

hope this helps u

pls mark as brainliest .-.

Answer:

See explanation

Explanation:

First voltaic cell;

Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)

Anode;

Zinc

Cathode;

Copper

Oxidation half equation;

Zn(s)------> Zn^2+(aq) + 2e

Reduction half equation;

Cu^2+(aq) +2e -----> Cu(s)

Overall; Zn(s) + Cu^2+(aq) -----> Zn^2+(aq) + Cu(s)

E°cell = 0.34 -(-0.76) =1.1 V

Second voltaic cell;

Zn(s)|Zn^2+(aq)||Fe^2+(aq)|Fe(s)

Anode;

Zinc

Cathode;

Iron

Oxidation half equation;

Zn(s)------> Zn^2+(aq) + 2e

Reduction half equation;

Fe^2+(aq) +2e -----> Fe(s)

Overall; Zn(s) + Fe^2+(aq) -----> Zn^2+(aq) + Fe(s)

E°cell = (-0.44) -(-0.76) = 0.32 V

Third voltaic cell;

Fe(s)|Fe^2+(aq)||Cu^2+(aq)|Cu(s)

Anode;

Iron

Cathode;

Copper

Oxidation half equation;

Fe(s)------> Fe^2+(aq) + 2e

Reduction half equation;

Cu^2+(aq) +2e -----> Cu(s)

Overall; Fe(s) + Cu^2+(aq) -----> Fe^2+(aq) + Cu(s)

E°cell = 0.34 -(-0.44) = 0.78 V

Fourth voltaic cell

Cu(s)|Cu^2+(aq)||I2(aq)|C(s)|I^-(aq)

Anode;

Copper

Cathode;

Graphite rod

Oxidation half equation;

Cu(s)------> Cu^2+(aq) + 2e

Reduction half equation;

I2(aq) +2e -----> 2I^-(aq)

Overall; Cu(s) + I2(aq) -----> Cu^2+(aq) + 2I^-(aq)

E°cell = 0.54 -0.34 = 0.20 V

Answer:

d. the conjugate base.

Explanation:

The general reaction of a weak acid, HA, with a strong base YOH, is:

HA + YOH → A⁻ + H₂O + Y⁻

Where A⁻ is the conjugate base of the weak acid and Y⁻ usually is a strong electrolyte.

That means after he complete reaction you don't have weak acid nor strong base, just conjugate base that will be in equilibrium with water, thus (Strong electrolyte doesn't change pH:

A⁻ + H₂O ⇄ HA + OH⁻

As the equilibrium is producing OH⁻, the pH of the solution is being affected for the conjugate base

Right option:

Answer:

5.74M or 5.74 mol/L (to 3 sign. fig.)

Explanation:

The molar mass of NaNO3 is 85g/mol, which means that:

1 mole of NaNO3 - 85g

? moles - 122.0g

= 122/85 = 1.44 moles

Concentration in mol/L = no. of moles (moles) ÷ volume (L)

[tex]\frac{1.44}{0.250}[/tex] = 5.74M or 5.74 mol/L (to 3 sign. fig.)

I hope the steps are clear and easy to follow.

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

The amount of heat to absorb is 6,261 J.

Heat required to raise the temperature of ice from -20 °C to 0 °C.

The formula for specific heat is used to calculate the amount of heat

Q = c * m * ΔT

Where,

Q =heat exchanged by a body,

m= mass of the body

c= specific heat

ΔT= change in temperature

Given:

m= 10 g,

specific heat of the ice= 2.1

ΔT=0 C - (-20 C)= 20 C

On substituting the values:

Q= 10 g*2.1  *20 C

Q=420 J

Heat required to convert 0 °C ice to 0 °C water.

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

Heat required to raise the temperature of water from 0 °C to 60 °C

m= 10 g,

Specific heat of the water= 4.18  

ΔT=60 C - (0 C)= 60 C

On substituting:

Q= 10 g*4.18  *60 C

Q=2,508 J

Thus, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

Find more information about Specific heat here:

brainly.com/question/13439286

Answer:

\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}  

Explanation:

0.030 cm³ × ? = x m³

You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.

For example, you know that centi means "× 10⁻²", so  

1 cm = 10⁻² m

If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).

If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.

So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.

We choose the former because it has the desired units on top.

The "cm" is cubed, so we must cube the conversion factor.

The calculation becomes

[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]

Answer:

2HClO(aq) + Ba(OH)₂(aq) →  Ba(ClO)₂(aq) + 2H₂O(l)

Explanation:

The reaction corresponds to a neutralization reaction between an acid and a base, as follows:

2HClO(aq) + Ba(OH)₂(aq)  →  Ba(ClO)₂(aq) + 2H₂O(l)            

From the equation above we have that the acid HClO reacts with the base Ba(OH)₂ to obtain a salt Ba(ClO)₂ and water.

In the balanced reaction, we have that 2 moles of HClO react with 1 mol of Ba(OH)₂ to produce 1 mol of Ba(ClO)₂ and 2 moles of water.

I hope it helps you!    

Answer:

[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}rightarrow[/tex]

Explanation:

Hello,

In this case, given that the sulfuric acid is a diprotic acid (two hydrogen ions) we can identify two ionization reactions, the first one, showing up the dissociation of the first hydrogen to yield hydrogen sulfate ions and the second one, showing up the dissociation of the hydrogen sulfate ions to hydrogen ions and sulfate ions by separated as shown below:

[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}[/tex]

They are have one-sensed arrow, since sulfuric acid is a strong acid.

Regards.

The equations that represent the first and second ionization steps for sulfuric acid in water are H₂SO₄→HSO₄+H⁺ & HSO₄⁻→SO₄⁻+H⁺ respectively.

Ionization reactions are those reactions in which atom or molecule will convert into ion by bearing a positive or negative charge on itself.

In water in the following way ionization of sulphuric acid takes place:

        H₂SO₄ → HSO₄⁻ + H⁺

        HSO₄⁻ → SO₄⁻ + H⁺

Hence, two steps are shown above.

To know more about ionization reaction, visit the below link:
https://brainly.com/question/1445179

Answer:

The boy must swim at an angle of 59°northwest to get to a position directly opposite his starting point.

Explanation:

To get to a point directly opposite his starting point, the boy must travel at an angle x, in a direction northwest of his starting point. The speed of the boy and the speed of the river current forms a right-angled triangle with an an opposite side of 1.8 m/a and a hypotenuse of 2.1 m/a having an angle x.

Sin x = opp/ hyp

Sin x = 1.8/2.1

x = sin⁻¹ (1.8/2.10

x = 58.99

x = 59°

Therefore, the boy must swim at an angle of 59° in the northwesterly direction to get to a position directly opposite his starting point.

Answer:

0.44

Explanation:

We know that the pH of any acid solution is given by the negative logarithm of its hydrogen ion concentration. Hence, if I can obtain the hydrogen ion concentration of any acid, I can obtain its pH.

For the acid, HNO3, [H^+] = [NO3^-]= 0.36 M

pH= -log [H^+]

pH= - log[0.36]

pH= 0.44

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm  ; Final Volume = 1.8 L ; Final pressure = 0.73 atm  

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T[tex]_2[/tex] ),

P[tex]_1[/tex]V[tex]_1[/tex] / T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / T[tex]_2[/tex],

T[tex]_2[/tex] = P[tex]_2[/tex]V[tex]_2[/tex]T[tex]_1[/tex] / P[tex]_1[/tex]V[tex]_1[/tex],

T[tex]_2[/tex] = 0.73 atm [tex]*[/tex] 1.8 L [tex]*[/tex] 298.15 K / 1 atm [tex]*[/tex] 1.2 L = ( 0.73 [tex]*[/tex] 1.8 [tex]*[/tex] 298.15 / 1 [tex]*[/tex] 1.2 ) K = 326.47425 K,

T[tex]_2[/tex] = 326.47425 K = 53.32425 C

Answer:

A. NaHCO₃

Explanation:

NaHCO₃ ⇒ NaOH + H₂CO₃

NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.

Hope that helps.

Answer:

[tex]\huge\boxed{Option \ 2}[/tex]

Explanation:

A reaction is spontaneous at all temperatures by the following combinations:

=> A negative enthalpy change ( [tex]\triangle H < 0[/tex] )

=> A positive entropy change ( [tex]\triangle S > 0[/tex] )

See the attached file for more better understanding!

from Gibbs Equation, [tex] \Delta G = \Delta H - T\Delta S [/tex]

reaction is spontaneous if $\Delta G$ is negative.

so, first option is not valid at high temperature, ($-h+ts$)

second, is always a spontaneous reaction, ($-h-ts$)

third, is never spontaneous ($+h+ts$)

4th is similar to second, spontaneous at higher temperatures ($+h-ts$)

Answer:

The answer is o2

Explanation:

I took the test

Answer:

7.38 g/cm³ is the density of the metal

Explanation:

In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).

To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.

Volume = a³

a = √8×r

(r = 198x10⁻¹²m)

a = 5.6x10⁻¹⁰ m

Volume = 1.756x10⁻²⁸ m³

1m = 100cm → 1m³ = (100cm)³:

1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =

There are 4 atoms of gold:

4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold

As 1 mole weighs 195.08g:

6.64x10⁻²⁴ moles of gold × (195.08g / mol) =

1.296x10⁻²¹g / 1.756x10⁻²² cm³ =

The density of the metal is 7.40 g/cm³

In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.

It is expressed by using the formula:

[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]

where;

For face-centered cubic FFC;

The edge length  [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]

[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]

[tex]\mathbf{a =560.0285 \ pm }[/tex]

a = 5.60 × 10⁻⁸ cm

Replacing it into the previous equation, we have:

[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]

[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]

Learn more about face-centered cubic arrangement here:

https://brainly.com/question/14786352?referrer=searchResults

Answer:

HBr, CH3C(OH)2 and CH3COOH, Br-

Explanation:

The conjugate acid-base pairs acid reacts with base to form a conjugate acid and conjugate base.

Conjugate acid is formed when a bases receives a proton  (H+) and a conjugate base is formed when an acid losses a proton (H+).

From the given equation:

HBr, CH3C(OH)2 and CH3COOH, Br- are conjugate acid-base pair, where HBr is an acid and CH3C(OH)2 is a conjugate acid while CH3COOH and Br- is the conjugate base.

Answer:

The 3rd one

Explanation:

Answer:

whats the ph  ofpoh=9.78

Explanation:

Answer:

See explanation

Explanation:

When we talk about electrophilic substitution, we are talking about a substitution reaction in which the attacking agent is an electrophile. The electrophile attacks an electron rich area of a compound during the reaction.

The five membered furan ring is aromatic just as benzene. This aromatic structure is maintained during electrophilic substitution reaction. The attack of the electrophile generates a resonance stabilized intermediate whose canonical structures have been shown in the image attached.

Answer:

Empirical formulae is NO2

Molecular Formulae is NO2

Explanation:

Answer:

1. 36

2. Two

Explanation:

The Lewis structure shows the valence electrons present in a compound. Usually the valence electrons are shown as dot structures around the symbol of the elements involved in the compound.

For a compound AB4 where B is more electronegative than A and A has 8 electrons in its valence shell, there will be thirty six valence electrons on the outermost shell of the molecule.

There are six electron pair domains present in the molecule, four bond pairs and two lone pairs. The molecule is in a square planar geometry.

Answer: a- 36 valence electrons

b- 14 lone pairs

Explanation:

Valence is equal to A + 4B = 8 + 4(7)

With 4 bonds between A and the 4 B, that is 36 valence minus 8 electrons in those pairs = 28. 28 is 14 lone pairs.

Answer:

123.96 g Na₂O

Explanation:

4 Na  +  O₂  ⇒  2 Na₂O

You first need to find the limiting reagent.  Convert the reactants to moles and see which produces the least amount of product using the mole ratios in the chemical equation.

(91.96 g Na)/(22.99 g/mol Na) = 4 mol Na

(4 mol Na) × (2 mol Na₂O/4 mol Na) = 2 mol Na₂O

(32.0 g O₂)/(32.0 g/mol) = 1 mol O₂

(1 mol O₂) × (2 mol Na₂O/1 mol O₂) = 2 mol Na₂O

Since they both produce the same amount of product, you don't need to pick a limiting reagent.  Now, convert moles of Na₂O to grams.

(2 mol Na₂O) × (61.98 g/mol Na₂O) = 123.96 g Na₂O

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = 4.74

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

Answer:

ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 )

Explanation:

Isomers are compounds that have the same molecular formula but different structural formulas. Hence any pair of compounds that can be represented by exactly the same molecular formula are isomers of each other.

If we look at the pair of compounds; ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ), one compound has molecular formula, C2H7ON while the other has a molecular formula, C2H5ON, hence they are not isomers of each other.

Answer:

(a )people separate mixtures in order to ger a specific substance that they need.

Answer:

H₂CO₃   H₂O  ⇄  HCO₃⁻  +  H₃O⁺          Ka1

HCO₃⁻  +  H₂O  ⇄  CO₃⁻²  +  H₃O⁺        Ka2

CO₃⁻²  +  H₂O  ⇄  HCO₃⁻   +  OH⁻       Kb1

HCO₃⁻  +  H₂O  ⇄   H₂CO₃  +  OH⁻     Kb2

Explanation:

Formula for carbonic acid is: H₂CO₃

It is a dyprotic acid, because it can release two protons. We can also mention that is a weak one. The equilibrums are:

H₂CO₃   H₂O  ⇄  HCO₃⁻  +  H₃O⁺          Ka1

HCO₃⁻  +  H₂O  ⇄  CO₃⁻²  +  H₃O⁺        Ka2

When the conjugate strong bases, carbonate and bicarbonate take a proton from water, the reactions are:

CO₃⁻²  +  H₂O  ⇄  HCO₃⁻   +  OH⁻       Kb1

HCO₃⁻  +  H₂O  ⇄   H₂CO₃  +  OH⁻     Kb2

Notice, that bicarbonate anion can release or take a proton to/from water. This is called amphoteric,

Answer:

Check the Attachment.

Half-reactions:

Anode: (OXIDATION) Ni --> Ni2+ + 2e-

Cathode: (REDUCTION) Cu2+ +2e- --> Cu

Overall reaction: Ni + Cu2+ --> Ni2+ + Cu

Explanation:

Overall, reaction is basically Anode + Cathode, where electrons on both sides cancel out  (if not, you need to multiply the equation in a way you can cancel them out).

Hope this helps.

Answer:

The best reducing agent is Zn(s)

Explanation:

A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound  with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:

Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V

Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

Ni²⁺(aq)|Ni(s) ⇒‑0.25 V

Cu²⁺(aq)|Cu(s) ⇒ +0.34 V

Ag⁺(aq)|Ag(s) ⇒ +0.80 V

The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).