Substance B is hard, does not conduct electricity, and melts at 1200 °C. Substance B is likely a(n):
(a) ionic solid
(b) metallic solid
(c) molecular solid
(d) covalent network solid

Answer:

0 degree C

Explanation:

0 degree C = 32 degree F

0 degree F = -17.7778 degree C

Answer:

0 degrees C would be warmer

Explanation:

Answer:

Explanation:

given mass = 697 g

molar mass = mass of H × mass of N × 3 × mass of O

= 1 × 14 × 3 × 16 = 672 g

no. of moles = given mass / molar mass = 697 / 672 = 1.03 moles

Hope this helps

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Reactants take 504.87 yr to reach 12.5% of their original value in first-order decomposition reaction.

[tex]A_{t} =A_{0} e^{-kt}[/tex]....(1)

Here,  [tex]A_{t}[/tex] is the final concentration, t is the time,  [tex]A_{0}[/tex] is the initial concentration, and k is the rate constant.

Given:-

[tex]A_{t} =0.125A_{0}[/tex]

k= [tex]0.00140yr^{-1}[/tex]

Substitute the above value in equation (1) as follows:-

[tex]0.125A_{0} =A_{0} e^{-kt} \\0.125A_{0} =A_{0} e^{-k\times0.00140 yr^{-1} }\\ln(0.125)/(-0.00140)=t\\t=504.87 year[/tex]

So, 504.87 yr does it take for the reactant to reach 12.5% of its original value.

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Answer:

Overall charge = -3

Explanation:

The atom of every chemical element is its smallest indivisible part. However, this atom further consists of subatomic particles namely: proton, electron, and neutron. The proton and electron are the positively charged and negatively charged particle respectively.

In a neutral atom, the amount of proton and electrons in that atom equates. However, the amount of electron and proton present in an atom determines the charge of that atom. For example, in this question, an atom is said to contain 17 protons, 17 neutrons and 20 electrons.

Since the negatively charged electrons (20) are more than the positively charged protons (17) by 3, hence, the net charge is -3.

Answer: put 45 mm

Explanation:

Answer:

D) 2( 6.02 ×10^23)

Explanation:

using n=m

M

where n = moles

m=mass of substance

M=molar mass .

n( sulfur ) =64

32

= 2

using N =nL

where N is number of entities

n = moles

L= 6.02×20^23

N( of atoms of sulfur) =2(6.02×10^23)

32(6.02 x 10²³) number of atoms are in 64 g sulfur (S) is:

C. 32(6.02 x 10²³) atoms

To determine the number of atoms in 64 grams of sulfur (S), you need to use Avogadro's number, which is approximately 6.02 x 10²³ atoms/mol. This number represents the number of atoms or molecules in one mole of a substance.

1. Find the molar mass of sulfur (S):

The molar mass of sulfur (S) is the mass of one mole of sulfur atoms, and you can find it on the periodic table. The molar mass of sulfur is approximately 32.06 g/mol.

2. Calculate the number of moles of sulfur:

To calculate the number of moles, divide the given mass (64 g) by the molar mass of sulfur (32.06 g/mol):

Number of moles = Mass of sulfur / Molar mass of sulfur

Number of moles = 64 g / 32.06 g/mol

Number of moles = 1.997 moles

3. Use Avogadro's number to find the number of atoms:

Now, multiply the number of moles by Avogadro's number:

Number of atoms = Number of moles × Avogadro's number

Number of atoms = 1.997 moles × 6.02 x 10²³ atoms/mol

Number of atoms = 1.202 x 10²⁴ atoms

So, there are approximately 1.202 x 10²⁴ atoms in 64 grams of sulfur (S). The closest option to this value is:

C. 32(6.02 x 10²³) atoms (which is equivalent to 1.924 x 10²⁴ atoms)

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Answer:

15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.

Explanation:

To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid

4.25 = 3.75 + log [A⁻] / [HA]

0.5 = log [A⁻] / [HA]

3.162 = [A⁻] / [HA] (1)

As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:

0.10M  * 20x10⁻³L =

2x10⁻³moles = [A⁻] + [HA] (2)

Replacing (2) in (1):

3.162 = 2x10⁻³moles - [HA] / [HA]

3.162 [HA] = 2x10⁻³moles - [HA]

4.162[HA] = 2x10⁻³moles

[HA] = 4.805x10⁻⁴ moles

[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles

That means, to create the buffer you must add:

[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =

[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =

Answer:

pOH = 11

pH = 3

Explanation:

Given data:

Hydrogen ion concentration = 10⁻³ mol/L

pOH of solution = ?

pH of solution = ?

Solution:

pH = -log[H⁺]

Now we will put the values.

pH = -log[ 10⁻³]

pH = -(-3) log 10

pH = 3

pOH of solution:

pOH + pH = 14

pOH = 14 - pH

pOH = 14 - 3

pOH = 11

Answer:

Several extractions is more effective than a single extraction.

Explanation:

When extraction is carried out multiple times, for instance, in this case, the extraction was carried out with three portions of the solvent rather than with just a single portion of the solvent, the amount of material left in the residue will be lower, because the extraction is more complete.

Several extractions with smaller volumes of solvent are more effective than a single extraction with a large volume of solvent.

Answer:

Solar Cell

Explanation:

Claim 1: The sun (solar cells) is the best energy source for the rescue team.

Answer:

False

Explanation:

From the question given above, the following data were obtained:

Electron number = 6

Proton number = 6

Neutron number = 6

From the question given above, we can see clearly that the model has the same number of protons and electrons. Hence the model is not an ion.

This can further be explained if we determine the charge.

Electron number = 6

Proton number = 6

Charge =?

Charge = Proton – Electron

Charge = 6 – 6

Charge = 0

Since the charge is zero, the model is not an ion.

Answer:

pH = 9.26

Explanation:

Let's make the reaction of this weak base:

C₅H₅N  +  H₂O  ⇄    C₅H₅NH⁺  +  OH⁻     Kb

We do not know the Kb, but we can discover it from pKb

Kb = 10^⁻pKb  →  10⁻⁸'⁷⁷ = 1.70×10⁻⁹

So, let's make the expression for Kb

Kb = [OH⁻] . [C₅H₅NH⁺] / [C₅H₅N]

In the equilibrum, we would have  0.20 moles from the begining - x (react)

So we would proudce x moles of OH⁻ and x moles of C₅H₅NH⁺.

In conclussion:

1.70×10⁻⁹ = x . x / (0.20 - x)

To avoid the quadratic formula we can miss the x from the substraction

1.70×10⁻⁹ = x² / 0.20

1.70×10⁻⁹ . 0.20 = x²

We avoid the x from the (0.20-x), because the base is so concentrated, and Kb is very small.

x = √(1.70×10⁻⁹ . 0.20) → 1.84×10⁻⁵

That's the value of [OH⁻], so we can calculate pOH

- log [OH⁻] = pOH → - log 1.84×10⁻⁵ = 4.73

Then pH = 14 - pOH

14 - 4.73 = 9.26

The pH of a 0.20 M aqueous solution of pyridine is approximately 9.26.

Given,

pKb = 8.77

Concentration = 0.20M

The pKb is related to the pKa by the equation: pKa + pKb = 14.

pKa = 14 - 8.77 = 5.23.

Since pyridine is a base, it can be treated as the conjugate base of its corresponding acid. The acid is the pyridinium cation, which is formed by the protonation of pyridine.

Now, the equilibrium between the pyridine (base) and the pyridinium ion (acid):

C₅H₅N + H₂O ⇌ C₅H₅NH⁺ + OH⁻

The equilibrium constant expression for this reaction can be expressed as:

Kb = [tex]\frac{([C_5H_5NH^+][OH^-])}{[C_5H_5N]}[/tex]

At equilibrium, the concentration of OH⁻ will be equal to the concentration of [C₅H₅NH⁺], as pyridine is a weak base and will not significantly dissociate in water.

Kb = [tex]\frac{[OH^-]^2}{[C_5H_5N]}[/tex]

Kb = [tex]10^{(-8.77)[/tex]

Kb = [tex]\frac{[OH^-]^2}{[C_5H_5N]}[/tex]

[tex]10^{(-8.77)[/tex] = [tex]\frac{[OH^-]^2}{(0.20 M)}[/tex]

[OH⁻]² = [tex]10^{(-8.77)[/tex] ₓ 0.20 M

[OH⁻] = [tex]\sqrt{10^{-8.77} \times 0.20M}[/tex]

[OH⁻] ≈ 1.42 x 10⁻⁵ M.

The pOH (the negative logarithm of the hydroxide ion concentration):

pOH = [tex]-log_{10}([OH^-])[/tex]

pOH = [tex]-log_{10}[/tex](1.42 x 10⁻⁵)

pOH ≈ 4.85

Since pH + pOH = 14, the pH:

pH = 14 - pOH

pH = 14 - 4.85

pH ≈ 9.26

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Answer:

1.0 M

Explanation:

Reaction equation;

KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)

Concentration of acid CA = ?

Concentration of base CB = 1.0 M

Volume of base VB = 25.60 - 0.50 = 25.1 ml

Volume of acid VB =  25.0 ml

Number of moles of acid NA = 1

Number of moles of base NB =2

CAVA/CBVB =NA/NB

CAVANB = CBVBNA

CA = CBVBNA/VANB

CA = 1 * 25.1 * 1/25.0 *1

CA = 1.0 M

The answer is a                                  bevasue it then becomes a chemical compound

Answer:

a

Explanation:

this would result in a compound and compounds are chemical changes so i think im right....

Answer:

The frequency is  [tex]f =  2,01 * 10^{15} \  Hz  [/tex]

Explanation:

From the question we are told that

   The energy required to ionize boron is [tex]E_b  =  801 KJ/mol[/tex]

Generally the ionization energy of boron pre atom is mathematically represented as

     [tex]E_a  =  \frac{E_b}{N_A}[/tex]

Here  [tex]N_A[/tex] is the Avogadro's constant with value [tex]N_A  =  6.022*10^{23}[/tex]

So

      [tex]E_a  =  \frac{801}{6.022*10^{23}}[/tex]

=>     [tex]E_a  =  1.330 *10^{-18} \  J/atom [/tex]

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

       [tex]E =  hf  =  E_a[/tex]

=>     [tex]  hf  =  E_a[/tex]

Here h is the Planks constant with value [tex]h = 6.626 *10^{-34}[/tex]

So

       [tex]f =  \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}[/tex]

=>      [tex]f =  2,01 * 10^{15} \  Hz  [/tex]

Answer:

A compound

Explanation:the mixture can vary

Answer:

mass = 508 g

Explanation:

Given data:

Number of moles of Cu = 8 mol

Mass in gram = ?

Solution:

Number of moles = mass/molar mass

Molar mass of Cu is 63.5 g/mol.

Now we will put the values in formula.

8 mol = mass /63.5 g/mol

mass = 63.5 g/mol ×8 mol

mass = 508 g

Answer:

Covalent bonds

Explanation:

i'm pretty sure it's fusion but i may be wrong.

Answer:

Flame tests use simple equipment, making them ideal for fieldwork. Geologists use the flame test to identify the presence of metals. Forensic scientists can use flame tests at crime scenes for quick analysis of elements present. Miners use the test for analysis of samples, particularly when prospecting. Flame tests provide a good teaching tool for chemistry students learning about emission spectra.

Unlike more sophisticated spectrographic equipment, a flame test requires only a gas burner, a hydrochloric acid solution and nichrome wire to hold the sample. The process is simple: dip the nichrome wire in the acid solution and hold it in the flame to remove any impurities, then affix the sample and hold it in the flame. The emitted colors show what metal ions the sample contains. For example, Copper emits a deep blue, Sodium bright orange and Lead a grey-white color. A table of elements and their characteristic colors helps with identification.

Explanation:

Answer:

[tex]x_{CO_2}^{out} =0.25[/tex]

Explanation:

Hello.

In this case, for the reactive scheme, it is very convenient to write each species' mole balance as shown below:

[tex]CH_4:f_{CH_4}^{out}=f_{CH_4}^{in}-\epsilon \\\\O_2:f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon\\\\CO_2:f_{CO_2}^{out}=\epsilon\\\\H_2O:f_{H_2O}^{out}=2\epsilon[/tex]

Whereas [tex]\epsilon[/tex] accounts for the reaction extent. However, as all the methane is consumed, from the methane balance:

[tex]0=f_{CH_4}^{in}-\epsilon \\\\\epsilon=30gmol/s[/tex]

Thus, we can compute the rest of the outlet mole flows since not all the oxygen is consumed as it is in excess:

[tex]f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon=75gmol/s-2*30gmol/s=15gmol/s\\\\f_{CO_2}^{out}=15gmol/s\\\\f_{H_2O}^{out}=2*15gmol/s=30gmol/s[/tex]

It means that the mole fraction of carbon dioxide in that output is:

[tex]x_{CO_2}^{out}=\frac{15}{15+15+30} =0.25[/tex]

Best regards.

Answer:

The equilibrium concentration of IBr is 0.0360 M or 0.1032 M

The equilibrium concentration of IBr is 0.752 mol

Explanation:

Mathematically, concentration = number of moles/volume

For Iodine molecule, concentration = 0.2/5 = 0.04 M

For the bromine molecule, concentration = 0.1/5 = 0.02 M

We need to set up an ICE table ( Initial, change, equilibrium)  to get the concentrations of the molecules at equilibrium.

Please check this table in the attachment.

From the table, we can see that the equilibrium concentration of Iodine is (0.04-x), that of Bromine is (0.02-x) while that of IBr is 2x

Now, we need to get the equilibrium constant  using the equilibrium concentration of the molecules

With the given chemical reaction, we can represent the equilibrium constant in terms of pressure as follows;

Kc = [IBr]^2/[Br2][I2]

Thus;

Kc = (2x)^2/(0.04-x)(0.02-x)

Kc = 4x^2/(0.04-x)(0.02-x)

From the question, Kc = 29 , so substitute this;

29 = 4x^2/(x^2-0.06x+0.008)

Cross multiply

29((x^2-0.06x+0.008) = 4x^2

29x^2 - 1.74x +0.0232 = 4x^2

Thus;

25x^2 - 1.74x + 0.0232 = 0

Solving this quadratic equation, we get two values for x which are x = 0.0516 and 0.0180

The equilibrium concentration is either (2 * 0.018) or (2 * 0.0516) which is 0.0360 or 0.1032

Answer:

1.88 × 10²⁴ atoms

Explanation:

Step 1: Given data

Mass of sulfur: 100 g

Step 2: Calculate the moles corresponding to 100 g of sulfur

The molar mass of sulfur is 32.07 g/mol. The moles corresponding to 100 g of sulfur are:

100 g × (1 mol/32.07 g) = 3.12 mol

Step 3: Calculate the number of atoms in 3.12 moles of sulfur

We will use Avogadro's number: there are 6.02 × 10²³ atoms of sulfur in 1 mole of sulfur.

3.12 mol × (6.02 × 10²³ atoms/1 mol) = 1.88 × 10²⁴ atoms

Answer:

ejeb094

Explanation:

nnb3neneie9eei rje

Answer:

19.3 g/cm3

Explanation:

A 10.0 cm3 sample of copper has a mass of 89.6 g. What is the density of copper? 19.3 g/cm3.

Answer:

height = 4.74 cm

density = 2.23 g/ cm³

Explanation:

Mass of metal = 0.650 kg (650 g)

Width = 0.136 m    

Length = 0.0451 m  

Volume of metal = 291 cm³

Height in cm = ?

density of metal =?

Solution:

Width = 0.136 m    (0.136 m×100 cm/1m = 13.6 cm)

Length = 0.0451 m  (0.0451 m×100 cm/1m = 4.51 cm)

First of all we will calculate the height:

Volume = height× width× length

291 cm³ =      h     × 13.6 cm × 4.51 cm

291 cm³ =      h     × 61.34 cm²

h = 291 cm³ / 61.34 cm²

h = 4.74 cm

Density:

d = m/v

d = 650 g/291 cm³

d = 2.23 g/ cm³

Answer:

Explanation:

To determine the molar volume of the gas according to the equation at stp;

1 mole = 22.4 dm³

2 moles = 44.8 dm³

To determine the mass of NaN₃ inflated according to the equation

2NaN₃ (where Na = 23g and N = 14) = (2 × 23) + 2(14 × 3)

= 130 g

Hence, if 130g of NaN₃ is required to inflate 44.8 dm³ airbag upon impact

what mass of NaN₃ is required to fully inflate the air bag upon impact;

130g ⇒ 44.8 dm³

? ⇒ 11.9 dm³ (dm³ is same as L)

? = 130 × 11.9/44.8

? = 34.5g

34.5g of NaN₃ is required to fully inflate 11.9 L of air bag upon impact

The answer is number 4