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Ivahew [28]
3 years ago
5

A displacement reaction takes place when calcium is placed in a solution of zinc chloride. Which substances will most likely be

produced by the reaction?
Ca + ZnCl2 → ?

A.
Zn and CaCl2
B.
CaZn and Cl2
C.
CaCl and ZnCl
D.
Ca, Zn, and Cl2
Chemistry
1 answer:
Karolina [17]3 years ago
5 0

Hi!


The correct option would be A.


This is because the displacement reaction would take place as follows

Ca + ZnCl2 --> CaCl2 + Zn


A displacement reaction is one in which a substitution occurs, as the more reactive element in the mixture replaces one that is less reactive.

In the electrochemical series, we find Ca higher than Zn, which is indicative of Ca being more reactive, and having the capacity to displace Zn to form a compound.


Option D would be incorrect as no such substitution occurs.

Option B would be incorrect because again, there is no substitution occurring, and also because two metals alone (Ca and Zn in our case) can never react to form a compound.

Option C would be incorrect because it is not possible because CaCl and ZnCl are forms that are too unstable to exist due to an overall positive charge.


Hope this helps!

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If 20 mL of gas is subjected to a temperature change from 10°C to 100°C and a pressure change from 1 atm to 10 atm,
Evgesh-ka [11]

Answer:

E. None of these

Explanation:

We know, By GAS laws,

PV = NRT, where p- pressure, v- volume, n- number of moles, R- gas constant ,and T- temperature

Now, In the question, the number of moles remains the same as the gas is the same. so n is constant so we can compare n before and after a temperature change.

\frac{P1V1}{RT1} = \frac{P2V2}{RT2}

where P1= 1 atm, P2 = 10 atm, V1= 20 mL, T1= 10°C and T2= 100°C

We don't have to worry about the standard units as they are present equally on both the sides and get cut, same goes for R( gas constant)

So putting values, we get

\frac{1*20}{R*10} = \frac{10*V2}{R*100}

Cutting, R on both sides and moving contents to the right so that only V2 is left on the left.

\frac{1*20*100}{10*10} = V2

∴ V2 = \frac{2000}{100}

∴ V2 = 20mL

3 0
1 year ago
Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec
Bess [88]

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0

Let energy change be E for 1 atom.

E=E_{\infty}-E_1=0-(-13.6  eV)=13.6 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol

E'=1,312.17 kJ/mol

The energy  required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ ,B^{4+} atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-340eV)=340 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 340eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol

E'=32,804.31 kJ/mol

The energy  required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ ,Li^{2+}atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol

E'=11,809.55 kJ/mol

The energy  required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ ,Mn^{24+}atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol

E'=820,107.88 kJ/mol

The energy  required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0
3 years ago
2. Problem: Is colored or black and white text more easily remembered? Hypothesis: If the text is colored, then the text is more
Varvara68 [4.7K]

Answer:

This question contains a lot of subsections. Please find the variables to each question below

Explanation:

Hypothesis is a testable explanation to a scientific question. It aims at proffering a testable solution to an observed problem. It usually goes in an IF, THEN format.

Dependent variable is the variable being measured in an experiment, while the independent variable is the variable that is manipulated or changed by the experimenter.

The control group is the group that does not receive the experimental treatment while constant is the variable that is kept unchanged for all groups in the experiment.

Based on this explanation, find the answers to the hypothesis, independent variable, dependent variable, constant, and control groups to each question.

2. Dependent variable: How easy it is to remember

Independent variable: Color of the text

Control: Group given Black and White text to read

Constants: Same text

3. Hypothesis: If the temperature is increased, THEN the speed of ants increases or move faster

Independent variable: TEMPERATURE

Dependent variable: SPEED OF ANTS

Control: Ants placed in room temperature

Constants: SAME KIND OF ANTS

4. Hypothesis: IF the pH of the soil is acidic or alkaline, THEN the growth of strawberries will be inhibited

Independent variable: pH of soil

Dependent variable: Growth of strawberries

Control: Strawberries grown in a Neutral pH soil

Constants: Same type of plants (strawberries)

5. Hypothesis: IF caffeine is taken, then the blood pressure will be increased or raised.

Independent variable: CAFFEINE

Dependent variable: BLOOD PRESSURE

Control: Group given no caffeine

Constants: Same amount of caffeine

6. Hypothesis: IF music is played, THEN the heart rate increases

Independent variable: MUSIC

Dependent variable: HEART RATE

Control: Groups placed where no music is played

Constants: Same type of music, same time music runs for etc.

7. Hypothesis: IF bleach is present/used, THEN bacteria will die

Independent variable: Bleach

Dependent variable: Death of bacteria

Control: Bacteria given no bleach

Constants: Same species of bacteria, Same type of bleach

3 0
3 years ago
What is the effect of this??
a_sh-v [17]

Answer:

Explanation:

The wavelength is the distance between two adjacent wavefronts. ... If the wave crosses to the new medium at an angle (not 90 degrees), the change ... When light enters a more optically dense medium, it is refracted closer to the normal. the same as the critical angle, light will travel along the boundary of the 2 mediums.

7 0
3 years ago
Read 2 more answers
Write the expression for the equilibrium constant Kp for the following reaction. (Enclose pressures in parentheses and do NOT wr
Liono4ka [1.6K]

Answer and Explanation:

For the following balanced reaction:

PCl₅(g) ↔ PCl₃(g) + Cl₂(g)

We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:

Kp=\frac{(P PCl_{3})(P Cl_{2})}{(P PCl_{5})}

Where PPCl₃ is the partial pressure of PCl₃ (reactant), PCl₂ is the partial pressure of Cl₂ (reactant) and PPCl₅ is the partial pressure of PCl₅ (product).

6 0
2 years ago
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