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4 years ago

9

Choose all the answers that apply. The elements in group 18 _____. are highly reactive are gases at room temperature usually do

not react with other elements are in a group called the noble gases are nonmetals

1 answer:

lubasha [3.4K]4 years ago

6 0

Answer:

Helium, Krypton, Neon, Argon, Radon, Oganesson and Xenon.

Explanation:

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What determines the color of a translucent object?<br> Please help due November 4.

Umnica [9.8K]

Answer: Color is a property of light that depends on wavelength. The color of a transparent object is determined by the wavelength of the light transmitted by it. An opaque object that reflects all wavelengths appears white; one that absorbs all wavelengths appears black.

Explanation:

6 0

3 years ago

How many grams of oxygen are required to react with 12.0 grams of octane(C8H18) in the combustion of octane in gasoline?

solniwko [45]

Answer : 42 grams of oxygen are required.

Explanation :

Step 1 : Write balanced chemical equation.

The combustion reaction of octane with oxygen can be written as,

$2 C_{8}H{18} (g) + 25 O_{2} (g) \rightarrow 16 CO_{2} (g) +18 H_{2}O (g)$

Step 2 : Find moles of octane.

The given mass of octane is 12 g

Molar mass of octane is 114.2 g/mol

The moles of octane are calculated as,

$moles =\frac{grams}{MolarMass} =\frac{12g}{114.2g/mol} = 0.105 mol$

We have 0.105 mol octane.

Step 3: Use mole ratio from balanced equation to find moles of O₂

The mole ratio of octane and O₂ is 2 : 25. Let us use this as a conversion factor.

$0.105 mol C_{8}H_{18} \times\frac{25mol O_{2}}{2molC_{8}H_{18}} = 1.31 mol$

We have 1.31 mol O₂.

Step 4 : Convert moles of O₂ to grams.

The molar mass of O₂ is 32 g/mol.

The grams of O₂ can be calculated as,

$1.31 mol O_{2} \times\frac{32g}{mol} = 42 grams$

42 grams of oxygen are required.

8 0

3 years ago

For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The

Alik [6]

Answer : The rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-2}M^{-2}s^{-1}$

Thus, the value of the rate constant 'k' for this reaction is $7.5\times 10^{-2}M^{-2}s^{-1}$

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1$

$\text{Rate}=4.25\times 10^{-2}Ms^{-1}$

Therefore, the rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

3 0

3 years ago

ArbitrLikvidat [17]

Answer:

c. 3.00 M HCl

Explanation:

From dilution formula

C1V1 = C2V2

C1=?, V1= 10.0ml, C= 1.5, V2= 20.0ml

Substitute and Simplify

C1×10= 1.5×20

C1= 3.00M

5 0

3 years ago

An aerosol can was found to burst at a pressure of about 24 atm and 85 c. What was the pressure inside the can at normal room te

baherus [9]

Answer:

Drd

Explanation:

6 0

3 years ago