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kondaur [170]
3 years ago
10

Calculate the density, in grams per liter, of carbon dioxide (CO2) gas at STP.

Chemistry
2 answers:
Simora [160]3 years ago
8 0

Answer: 1.96g/L

Explanation:

1mole of CO2 contains 22.4L at stp.

1mole of CO2 = 12 + ( 2x16) = 12 + 32 = 44g

Density = Mass /volume

Density = 44g /22.4L

Density = 1.96g/L

viva [34]3 years ago
5 0

Answer:

Density of CO₂ at STP = 1.96 g/L

Explanation:

1 mol of any gas at STP occupies 22.4L of volume.

The rule for the Ideal gases.

Assume 1 mol of CO₂, we know that has 44 grams of mass.

Density = mass / volume

44 g/ 22.4L = 1.96 g/L

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AlCl3 and NaOH <br> Balanced Equation<br> Total Ionic Equation<br> Net Ionic Equation
Sindrei [870]

Answer:

Here's what I get  

Explanation:

1. In dilute NaOH

(a) Molecular equation

AlCl₃(aq) + 3NaOH(aq) ⟶ Al(OH)₃(s) + 3NaCl(aq)

(b) Ionic equation

You write the molecular formulas for solids, and you write the soluble ionic substances as ions.

Al³⁺(aq) + 3Cl⁻(aq) + 3Na⁺(aq) + 3OH⁻(aq) ⟶ Al(OH)₃(s) + 3Na⁺(aq) + 3Cl⁻(aq)

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

Al³⁺(aq) + <u>3Cl⁻(aq)</u> + <u>3Na⁺(aq</u>) + 3OH⁻(aq) ⟶ Al(OH)₃(s) + <u>3Na⁺(aq)</u> + <u>3Cl⁻(aq) </u>

The net ionic equation is

Al³⁺(aq) + 3OH⁻(aq) ⟶ Al(OH)₃(s)

2. In excess NaOH

The aluminium hydroxide reacts with excess hydroxide to form sodium tetrahydroxoaluminate(III).

(a) Molecular equation

AlCl₃(aq) + 4NaOH(aq) ⟶ NaAl(OH)₄(aq) + 3NaCl(aq)

(b)  Ionic equation

Al³⁺(aq) + 3Cl⁻(aq) + 4Na⁺(aq) + 4OH⁻(aq) ⟶ Na⁺Al(OH)₄⁻(aq) + 3Na⁺(aq) + 3Cl⁻(aq)

(c) Net ionic equation

Al³⁺(aq) + 4OH⁻(aq) ⟶ Al(OH)₄⁻(aq)  

7 0
3 years ago
Light from three different lasers (A, B, and C), each with a different wavelength, was shined on the same metal surface. Laser A
kenny6666 [7]

Explanation:

It is known that each metal has a different work function and it is just the energy required to remove the electron from the surface.

The relation between energy and work function is as follows.

             \Delta E = Q + K.E

where,   \Delta E = energy of photon

               Q = work function

             K.E = kinetic energy of electron

When the value of Q is large then the amount of energy required by the photon is also large and vice-versa.

And, when there is not sufficient energy to remove an electron from metal surface or energy is less than its work function then electron will not be removed.

On the other hand, when incoming photon will have enough energy then electron will be ejected.

In the given situation, light from laser A is not able to remove the electron from metal surface which means that energy of photon is less than work function Q.

As, both B and C are able to remove electron from the surface of metal. So, after the removal of electrons rest of the energy of photons is transferred to eject electrons and this energy is converted into kinetic energy.

As energy of B is greater than the energy of C and A has the lowest energy among all.

Therefore, order of energy from lowest to highest is as follows.

                              A < C < B

Now, the relation between energy and wavelength is as follows.

                   E = \frac{hc}{\lambda}

where,    \lambda = wavelength

So, energy is inversely proportional to wavelength. Therefore, increase in energy means decrease in wavelength.

Therefore, increasing order of wavelength is as follows.

                            B < C < A

8 0
3 years ago
Find the empirical formula of the following compounds:
Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

#SPJ4

8 0
1 year ago
Which element is being represented by the electron configuration below? [Ar] 4s2 3d3
Nina [5.8K]
Answer:
chromium (Cr)
4 0
3 years ago
What mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric
Gre4nikov [31]

 The mass of NaCl  formed  is 8.307  grams


<u><em> calculation</em></u>

step 1: write the equation  for reaction

Na₂CO₃  + 2HCl → 2 NaCl  +CO₂ +H₂O

Step 2: find the  moles of Na₂CO₃

moles = mass/molar mass

 The  molar mass of Na₂CO₃  is = (23 x2) + 12 + ( 16 x3) = 106 g/mol

moles  = 7.5 g/106 g/mol =0.071 moles

Step 3: use the  mole ratio to determine the  mole of NaCl

Na₂CO₃:NaCl  is  1:2  therefore the moles of NaCl =0.07  x2 =0.142 moles


Step 4:  calculate mass  of NaCl

mass= moles x molar mass

the molar  mass of NaCl= 23 +35.5 =58.5 g/mol

mass  = 0.142  moles x 58.5 g/mol =8.307  grams

8 0
3 years ago
Read 2 more answers
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