A cube of aluminum measures 3 cm on each side and weighs 81 g. What is its density? (HINT: find the volume of the block first)

D=

M=

V=

Answer:

3. NH3 is limiting

4. 90 g NO

Explanation:

3. To find the limiting reactant, you have to convert from moles of reactant to moles of product. In this case, because we need to know how many grams of NO in the second question, we will use NO as our product. We use the ratios found in the chemical formula.

[tex]3 mol NH3*\frac{4 molNO}{4 mol NH3} = 3molNO[/tex]

[tex]6 mol O2*\frac{4 molNO}{5 mol O2} = 4.8molNO[/tex]

Because NH3 makes less moles of NO, it is the limiting reactant.

4. Now we need the molar mass to convert the moles of NO into grams of NO. The molar mass of NO is 30.01g/mol. This is the equation.

[tex]3 mol NO*\frac{30.01g NO}{1 mol NO} = 90.03gNO[/tex]

Because there are is only one significant figure in the original numbers, we round to one significant figure = 90 g NO.

Answer:

Cave Crystal

Explanation:

I believe it's cave crystal because sand dunes, river deltas, and valleys are all examples of deposition.

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 .

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:

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:

I think option D

I hope my answer is helpful to you

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Answer:

Mass = 2304 g

Explanation:

Given data;

Number of atoms of oxygen = 8.65×10²⁵

Mass in gram = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

For example,

1 mole = 6.022 × 10²³ atoms

8.65×10²⁵ atoms of oxygen× 1mol  /6.022 × 10²³ atoms

1.44×10² mol

144 mol

Mass in grams of oxygen:

Mass = number of moles × molar mass

Mass = 144 mol × 16 g/mol

Mass = 2304 g

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

Explanation:

It represent Alcohol group (—OH)

Answer: put 45 mm

Explanation:

Answer:

The molecules in the liquid are moving past one another. The hydrogen end of one water molecule is attracted to the oxygen end of another but only for a short time because they are moving. The molecules in ice are further apart than in liquid water. The molecules in ice are in fixed positions but still vibrate.

Explanation:

-google

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:

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:

A compound

Explanation:the mixture can vary

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:

try dividing the grams and the volume and see what number matches up with your answer

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:

ejeb094

Explanation:

nnb3neneie9eei rje

Answer:

acidic because of electrical issues and the body of electrical equipment

Answer:

They show that the number of atoms of each  element is the same in the products and  reactants

Explanation:

Balanced chemical equations show the conservation of mass in that the number of atoms of each element is the same in the products and reactants.

Answer:

C (They show that the number of atoms of each element is the same in the products and reactants.)

For the next question: C (2H2(g) + O2(g) → 2H2O(g))

Explanation:

For Edge

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: [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]

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.