Give an example of a change that would be a chemical change but not a physical change.

Answer:

It is D !!

Explanation:

Just did test

Answer:

1.24 kJ is required to convert 14 g of liquid from 43.5°C to 128.2°C

Explanation:

This is a typical calorimetry problem:

We have to assume, no heat is lost to sourrounding.

First of all, we need to go from 43.5°C to 97.4°C, the boiling point.

Q = Ce . m . ΔT

We replace data, 1.18° J/g . 14 g . (97.4°C - 43.5°C)

Heat for the first stage is: 890.4 Joules

Now we have to change the state, and we need the ΔH. As we do not have latent heat, we can proceed like this:

1 mol release 30.1 kJ at vaporization.

We convert the mass to moles → 14 g.  1mol/ 67.44g = 0.207 mol

0.207 mol will release (0.207 . 30.1 kJ) = 6.25 kJ

Now, we are at gaseous phase.

Q = Ce . m . ΔT → 0.792 J/g°C . 14g . (128.2°C - 97.4°C)

Q = 341.5 Joules

To determine the amount of heat, we sum all the obtained values:

890.4 Joules + 6250 Joules + 341.5 Joules = 1238.2 J

We convert to kJ →  1238.2 J . 1kJ / 1000J = 1.24 kJ

The heat required to convert 14.0 g of an unknown liquid at 43.5 °C to gas at 128.2 °C is 7.48 kJ.

We want to calculate the heat required to convert 14.0 g of an unknown liquid at 43.5 °C to gas at 128.2 °C.

We can divide this process in 3 steps.

We will calculate the heat for this step (Q₁) using the following expression.

Q₁ = c(l) × m × ΔT

Q₁ = (1.18 J/g･°C) × 14.0 g × (97.4 °C - 43.5 °C) = 890 J = 0.890 kJ

where,

We will calculate the heat for this step (Q₂) using the following expression.

Q₂ = (m/M) × ΔHvap

Q₂ = [14.0 g/(67.44 g/mol)] × 30.1 kJ/mol = 6.25 kJ

where,

We will calculate the heat for this step (Q₃) using the following expression.

Q₃ = c(g) × m × ΔT

Q₃ = (0.792 J/g･°C) × 14.0 g × (128.2 °C - 97.4 °C) = 342 J = 0.342 kJ

where,

Q = Q₁ + Q₂ + Q₃ = 0.890 kJ + 6.25 kJ + 0.342 kJ = 7.48 kJ

The heat required to convert 14.0 g of an unknown liquid at 43.5 °C to gas at 128.2 °C is 7.48 kJ.

Learn more about heating curves here: https://brainly.com/question/10481356

Answer:

bulb or pump, meniscus, outside

Explanation:

In order to use a pipet, place a BULB OR PUMP at the top of the pipet. Use this object to fill the pipet such that the MENISCUS of the liquid is even with the volume line. Release the liquid, touching the tip of the pipet to the side of the container if necessary to release the last drop OUTSIDE the pipet tip.

Answer:

it is Glucose hope it helps

Answer:

Glucose and DNA

Explanation:

Complex macromolecules are the combination of more than two molecules which are large and complex. The major types of complex macromolecules consists of 4 types which are carbohydrates, lipids, proteins, and nucleic acids.

Glucose is one of the carbohydrates, while DNA is another form of nuclei acid.

Answer:

1. 0.097 s

2. 0.420 M

Explanation:

To solve both questions we'll use the formula:

[A]ₓ = [A]₀ - kt

Where [A]ₓ is the concentration of A at a given time; and [A]₀ is the initial concentration.

1) We input the data given by the problem:

0.167 M = 0.700 M - 5.48 M/s * t

And solve for t:

t = 0.097 s

2) We input the new data:

[A]ₓ = 0.500 M - 0.361 M/s * 0.220 s

and solve for [A]ₓ:

[A]ₓ = 0.420 M

Answer:

323.22 ml

Explanation:

Given that :

Diameter, d = 6.8cm

Height, h = 8.9cm

V = arh

Recall :

Volume, V = πr²h

Radius, r = diameter / 2 = 6.8 / 2 = 3.4cm

V = π * 3.4^2 * 8.9

V = 323.21961 cm³

Recall:

1ml = 1cm³

Hence,

323.21961 cm³ = 323.21961 ml

Volume = 323.22 ml

Answer:

MgCO₃

Explanation:

From the question given above, we obtained:

MgF₂ + Li₂CO₃ —> __ + 2LiF

The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:

MgF₂ (aq) —> Mg²⁺ + 2F¯

Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯

MgF₂ + Li₂CO₃ —>

Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:

MgF₂ + Li₂CO₃ —> __ + 2LiF

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Therefore, the missing part of the equation is MgCO₃

Answer:

Independent

Explanation:

Independent Variable is the volume of the object. Dependent Variable is the mass of the object. So it

Answer:

3.769022740695677

Explanation:

Answer:We get 1.81×1024 water molecules... Explanation: We assess the molar quantity of water in the usual way... Number of moles=massmolar mass.

brainliest plz

Answer:

lol

Explanation:

lol

Answer:

The Importance of D-Day

The D-Day invasion is significant in history for the role it played in World War II. It marked the turn of the tide for the control maintained by Nazi Germany; less than a year after the invasion, the Allies formally accepted Nazi Germany's surrender.

0.836 mol Al

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Stoichiometry

Step 1: Define

[RxN - Balanced] 2Al (s) + 3H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + 3H₂ (g)

[Given] 0.418 mol Al₂(SO₄)₃

[Solve] x mol Al

Step 2: Identify Conversions

[RxN] 2 mol Al (s) → 1 mol Al₂(SO₄)₃ (aq)

Step 3: Stoich

Step 4: Check

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

Since our final answer already has 3 sig figs, there is no need to round.

Answer:

HC₂H₃O₂ + NaHCO₃ —> NaC₂H₃O₂ + CO₂ + H₂O

The coefficients are: 1, 1, 1, 1, 1

Explanation:

_HC₂H₃O₂ + _NaHCO₃ —> _NaC₂H₃O₂ + _CO₂ + _H₂O

To balance an equation, we simply do a head count of the individual elements and ensure they are balanced on both side.

For the above equation, we shall balance it as :

HC₂H₃O₂ + NaHCO₃ —> NaC₂H₃O₂ + CO₂ + H₂O

Reactant:

H = 5

C = 3

O = 5

Na = 1

Product:

H = 5

C = 3

O = 5

Na = 1

From the above, we can see that each element is the same on both side of the equation. Thus the equation is already balanced

HC₂H₃O₂ + NaHCO₃ —> NaC₂H₃O₂ + CO₂ + H₂O

The coefficients are: 1, 1, 1, 1, 1

Answer:

6.78 × 10⁻³ L

Explanation:

Step 1: Write the balanced equation

Mg₃N₂(s) + 3 H₂O(g) ⇒ 3 MgO(s) + 2 NH₃(g)

Step 2: Calculate the moles corresponding to 10.2 mL (0.0102 L) of H₂O(g)

At STP, 1 mole of H₂O(g) has a volume of 22.4 L.

0.0102 L × 1 mol/22.4 L = 4.55 × 10⁻⁴ mol

Step 3: Calculate the moles of NH₃(g) formed from 4.55 × 10⁻⁴ moles of H₂O(g)

The molar ratio of H₂O to NH₃ is 3:2. The moles of NH₃ produced are 2/3 × 4.55 × 10⁻⁴ mol = 3.03 × 10⁻⁴ mol.

Step 4: Calculate the volume corresponding to 3.03 × 10⁻⁴ moles of NH₃

At STP, 1 mole of NH₃(g) has a volume of 22.4 L.

3.03 × 10⁻⁴ mol × 22.4 L/mol = 6.78 × 10⁻³ L

Answer:

0.81 moles H2

Explanation:

The equation of the reaction is;

Mg(s) + 2HCl(aq) ------>MgCl2(aq) + H2(g)

Number of moles of HCl reacted = concentration * volume

Number of moles of HCl reacted = 6.44 * 250/1000

Number of moles of HCl reacted = 1.61 moles of HCl

If 2 moles of HCl produces 1 mole of H2

1.61 moles of HCl produces 1.61 * 1/2 = 0.81 moles H2

Answer:  The mole fraction of the solvent in a solution that contains 2.51 g glycerol dissolved in 21.10 mL ethanol is 0.93

Explanation:

Given : Volume of ethanol (solvent) = 21.10 ml

density of ethanol (solvent)= 0.789 g/ml

Mass of ethanol (solvent) = [tex]0.789g/ml\times 21.10ml=16.6g[/tex]

Mass of glycerol (solute) = 2.51 g

Mole fraction of a component is the ratio of moles of that component to the total moles present.

moles of ethanol =[tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{16.6g}{46g/mol}=0.36mol[/tex]

moles of glycerol =[tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{2.51g}{92g/mol}=0.027mol[/tex]

mole fraction of ethanol (solvent) = [tex]\frac{\text {moles of ethanol}}{\text {moles of ethanol + moles of glycerol}}=\frac{0.36}{0.36+0.027}=0.93[/tex]

The mole fraction of the solvent in a solution that contains 2.51 g glycerol dissolved in 21.10 mL ethanol is 0.93

Answer:

It would get colder

Explanation:

The lattice energy is the energy involved in the disruption of interactions between the ions of the salt. In this case, we have: ΔHlat  = 350 kJ/mol > 0, so it is an endothermic process (the energy is absorbed).

The solvation energy is the energy involved in forming interactions between water molecules and the ions of the salt. In this case, we have: ΔHsolv  = 320 kJ/mol > 0, so it is an endothermic process (the energy is absorbed).

The dissolution process involve both processes: the disruption of ion-ion interactions of the salt and the solvation process. Thus, the enthalphy change (ΔHsol) in the preparation of the solution is calculated as the addition of the lattice energy and solvation energy:

ΔHsol= ΔHlat + ΔHsolv = 350 kJ/mol + 320 kJ/mol = 370 kJ/mol

370 kJ/mol > 0 ⇒ endothermic process

Since the preparation of the solution is an endothermic process, it will absorb energy from the surroundings, so the solution would get colder.

Answer:

Car B has more inertia than Car A

Explanation:

Given that,

Mass of car A = 1500 kg

Mass of car B = 2000 kg

Inertia is directly proportional to the mass of an object. Inertia is the measure of the mass of an object.

In this case, the mass of car B is more than that of car A, it means the inertia of car B is more than that of car A.

Hence, the correct option is (c) "Car B has more inertia than Car A".

Answer:

See explanation

Explanation:

The balanced reaction equation of the reaction between SiO2 and HF is shown below;

SiO2(s) + 4HF(aq) = SiF4(g) + 2H2O(l)

b)

2NaF(aq) + CaCl2(aq) --------> 2NaCl(aq) + CaF2(s)

The complete ionic equation is;

2 Na^+(aq) + 2F^-(aq) + Ca^+(aq) + 2 Cl^-(aq)------> 2 Na^+(aq) + 2 Cl^-(aq) + CaF2(s)

The net ionic equation;

Ca^+(aq) + 2F^-(aq) -------> CaF2(s)

Answer:

Calcium Hydroxide is an ionic compound

Answer:

Mg donates two electrons to O

Explanation:

Lewis dot notation uses dots and crosses to represent valence electrons on atoms.

Magnesium is a metal and would donate or lose electrons during bonding.

Oxygen is a non metal and would gain electrons during bonding.

The correct option is;

Mg donates two electrons to O

Answer:

they are a lot tougher and are expecting only greatness, sort of like theater critiques.

Explanation:

Answer:

[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2

Explanation:

2H2S(g)⇋2H2(g)+S2(g)2H2S(g)⇋2H2(g)+S2(g)

The equilibrium constant expression in terms of concentrations is:

Kc=[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2.

The equilibrium expression for the given reaction can be written in terms of equilibrium constant which is the ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.

Equilibrium is a state for a reversible reaction where, the rate of forward reaction is equal to the rate of backward reaction. The rate of a reaction is the rate of decrease in the concentration of reactants or the rate of increase in the concentration of the products.

The given reaction at equilibrium state is written as:

[tex]\rm 2H_{2}S (g)\leftrightharpoons 2H_{2} (g)+ S_{2}(g)[/tex]

The equilibrium constant Kb is ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.

[tex]Kb = \rm \frac{[H_{2}S]^{2}}{[H_{2}]^{2} [S_{2}]}[/tex]

The rate of the reaction will be r = Kb [H₂]² [S₂].

To find more on equilibrium constant, refer here:

https://brainly.com/question/15118952

#SPJ2

Answer:

Explanation

I am sorry but please give detailed question