A silver coin with a mass of 14.1g has a density of 10.5 grams

- Explain how ionic bonds, covalent bonds, and metallic bonds each form and how.

-Dominant- [34]

Answer: -Ionic bonds form when one atom provides electrons to another atom. Covalent Bonds: Covalent bonds form when two atom shares their valence electrons. Metallic Bonds: Metallic bonds form when a variable number of atoms share a variable number of electrons in a metal lattice.

-Covalent Bonds.

Covalent Compounds. Contain no metals and no ions. Covalent compounds contain nonmetals only.

Example:

Ionic Compounds. A metal with a non-metal. Doesn't use prefixes for naming. Name the metal and change the nonmetal ending to -ide.

Explanation: Ionic bonds form when a nonmetal and a metal exchange electrons, while covalent bonds form when electrons are shared between two nonmetals. An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.

6 0

1 year ago

A velocity-time graph shows how what changes over time?

finlep [7]

Answer:

A velocity-time graph shows how velocity changes over time. The sprinter's velocity increases for the first 4 seconds of the race, it remains constant for the next 3 seconds, and it decreases during the last 3 seconds after she crosses the finish line.

4 0

2 years ago

A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________

gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

R = gas constant, T = temperature, Mm = molar mass of the gas particles

From the question

R = 8,314 J / mol K

T = temperature

Mm = molar mass, kg / mol

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

$\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s$

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

M₁ = molar mass sulfur dioxide = 64

M₂ = molar mass nitrogen triodide = 395

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5$

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0

2 years ago

A very simple assumption for the specific heat of a crystalline solid is that each vibrational mode of the solid acts independen

MatroZZZ [7]

Answer:

4.7 kJ/kmol-K

Explanation:

Using the Debye model the specific heat capacity in kJ/kmol-K

c = 12π⁴Nk(T/θ)³/5

where N = avogadro's number = 6.02 × 10²³ mol⁻¹, k = 1.38 × 10⁻²³ JK⁻¹, T = room temperature = 298 K and θ = Debye temperature = 2219 K

Substituting these values into c we have

c = 12π⁴Nk(T/θ)³/5

= 12π⁴(6.02 × 10²³ mol⁻¹)(1.38 × 10⁻²³ JK⁻¹)(298 K/2219 K)³/5

= 9710.83(298 K/2219 K)³/5

= 1942.17(0.1343)³

= 4.704 J/mol-K

= 4.704 × 10⁻³ kJ/10⁻³ kmol-K

= 4.704 kJ/kmol-K

≅ 4.7 kJ/kmol-K

So, the specific heat of diamond in kJ/kmol-K is 4.7 kJ/kmol-K

7 0

2 years ago

3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con

KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

$m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}$

c) Half life for first-order kinetics is computed by:

$t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}} =53.3min$

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

$C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M$

Best regards.

6 0

3 years ago