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yKpoI14uk [10]
3 years ago
5

What is the basic building block of matter to a chemist

Chemistry
1 answer:
Harman [31]3 years ago
5 0
The basic building block of matter is the atom.
You might be interested in
What is the reactant(s) of the reaction below?
svet-max [94.6K]
Answer : C

Reason : The product is MgO which is Magnesium Oxide + CO2 which is carbon dioxide because of the 2
3 0
3 years ago
A compound has been shown to contain 14.5 percent hydrogen and 85.5 percent carbon by mass.
Sedaia [141]

Answer:

  • <em>Option d. Its empirical formula is CH</em><em>₂</em><em>.</em>

Explanation:

The percent composition of the compound allow you to calculate the empirical formula of the compound but is not enough to calculate either the molar mass or the molecular formula. So, since now you can discard options b. and c.

Telling that it is a hydrocarbon (option e.) is true but very vague compared with finding the empirical formula. So, you can also discard the option e.

The fact that the product has a triple bond cannot be concluded from the percent composition, you should find the molecular formula to assert whether it contains or not a triple bond. So, you could discard option a., which lets you only with choice d.

Let us find the empirical formula to be certain that it is CH₂.

1.  <u>First, assume a basis of 100 g of compound</u>:

  • H: 14.5% × 100 g = 14.0 g
  • C: 85.5% × 100 g = 85.5 g

2. <u>Divide each element by its atomic mass to find number of moles</u>:

  • H: 14.0 g / 1.008 g/mol = 14.38 mol
  • C: 85.5 g / 12.011 g/mol = 7.12 mol

3. <u>Divide both amounts by the smallest number, to find the mole ratio</u>:

  • H: 14.38 mol / 7.12 mol ≈ 2
  • C: 7.12 mol / 7.12 mol = 1.

Hence, the ratio is 2:1 and the empirical formula is CH₂.

7 0
3 years ago
Which statement is true according to the kinetic theory?
Vlad [161]

Answer:

<em>Molecules of different gases with the same mass and temperature always have the same average kinetic energy - E.</em>

5 0
3 years ago
Read 2 more answers
A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark causes
ziro4ka [17]

Answer:

PCO2  = 0.6 25 atm

PSO2  = 1.2 75 atm

PO2 = 0.6  atm

Explanation:

Step 1: Data given

Volume = 10.0 L

Temperature = 100.0 °C

Pressure = 3.10 °C

After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm

Step 2: The balanced equation

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Step 3: Name the reactants and products

a = CS2

b = O2 before reaction

c = CO2

d = SO2

e = nS O2 after reaction with n = the number of moles

Step 4: Calculate moles before reaction

PV = nRT

n = PV/(RT)

(na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

(na + nb) = 1.0124

Step 5: Calculate moles after reaction

PV = nRT

n = PV/(RT)

nc + nd + ne) = PV/(RT) = (2.50 atm)*(10.0L) / ((0.08206 Latm/moleK)*(373.15K))

(nc + nd + ne) = 0.816 moles

Step 6: Calculate mol fraction

For  1 mole CS2 we need 3 moles O2  to produce 1 mole of CO2 and 2 moles of SO2

moles O2 remaining = ne = nb - 3na

moles CO2 produced = nc = na

moles SO2 producted = nd = 2na

(nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

nb = 0.816

. (na + nb) = 1.0124

na = 1.0124 moles - 0.816 moles = 0.208

which leads to  

nc = na = 0.208

nd = 2na = 2*0.208 = 0.416

ne = 0.816 - 3*0.208 = 0.192

mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

mole fraction O2 = 0.192 /(0.208 + 0.416 + 0.192) = 0.24

Step 6: Calculate partial pressure

PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

PO2 = 0.24 * 2.50 atm = 0.6  atm

Step 7: Control results

now let's verify a couple of things

PV = nRT

P = nRT/V

before rxn

P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

after rxn

P = ((0.208 +0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm

8 0
3 years ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
3 years ago
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