2 2 6 2 6 2 10 3
1s 2s 2p 3s 3p 4s 3d 4p
=

Answer: [tex]Pb(NO_3)_2(aq)+2NaOH(aq)\rightarrow 2NaNO_3(aq)+Pb(OH)_2(s)[/tex]

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The balanced chemical equation for a precipitate forms when aqueous solutions of lead(II) nitrate and sodium hydroxide are combined is:

[tex]Pb(NO_3)_2(aq)+2NaOH(aq)\rightarrow 2NaNO_3(aq)+Pb(OH)_2(s)[/tex]

Explanation:

It represent Alcohol group (—OH)

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:

[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:

Solar Cell

Explanation:

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

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: put 45 mm

Explanation:

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:

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

plz mark as brainliest!!!!!!

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.

Find more information about first- order decomposition reaction here:-

brainly.com/question/20607444

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:

A compound

Explanation:the mixture can vary

Answer:

Covalent bonds

Explanation:

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

Answer:

+0.02 V

Explanation:

For the reaction to be spontaneous, iron must be the cathode and Mercury the anode

Given that the standard cell potential is obtained from;

E°=E° cathode - E° anode

Since;

E° cathode = -0.77V

E° anode = 0.79 V

E°cell = -0.77 -(-0.79)

E°cell = +0.02 V

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:

ejeb094

Explanation:

nnb3neneie9eei rje

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

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:

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

Answer:

Explanation:

Let n be number of half lives taken .

10 = 70 x ( 1/2 )ⁿ

1/7 = ( 1/2 )ⁿ

2ⁿ = 7

n ln2 = ln 7

n = ln7 / ln2

= 1.9459 / .693

= 2.8

So 2.8 half lives will be required .

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....

Complete Question

The complete question is shown in the first uploaded image

Answer:

So the math expression is  

             [tex]heat  =  \frac{ 35. 7  KJ *  1900 \ gram }{ 1 \ gram }[/tex]

Explanation:

From the question we are told that

  The heat released for 1 gram of reactant consumed is  [tex]H  =  37.5 \ KJ/g [/tex]

   The mass of reactant considered is  [tex]m =  1.9 \ kg  =  1900 \  g[/tex]

So  if

             [tex]37.5 \ KJ [/tex] is produced for  1 gram

Then

              x kJ is produced for  1900 g  

=>   [tex]x  =  \frac{ 35. 7  KJ *  1900 \ gram }{ 1 \ gram }[/tex]

So the heat released is  

       [tex]heat  =  \frac{ 35. 7  KJ *  1900 \ gram }{ 1 \ gram }[/tex]

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:

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:

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

The answer is number 4

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