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4 years ago

What is the best name for the following compound? CCl4

Chemistry

1 answer:

Nitella [24]4 years ago

6 0

Answer:

The answer is b. Carbon Tetrachloride

Explanation:

Let's discard the other options:

a. It couldn't be carbon chloride because there are four (4) chloride ions in the molecular formula. The molecular formula for the carbon chloride must be CCI. So, this answer is wrong

c. It could be monocarbon tetrachloride because there are 4 chloride ions and 1 carbon ion, but it's not necessary to use the prefix "mono" because there is just one carbon ion in the formula. In that way, when we mention carbon in the name, there is implicit that there is just one carbon atom in the formula. So, this answer could be ok, but isn't the best.

d. It couldn't be calcium chloride because there isn't any calcium ion or atom in the formula. The molecular formula for the calcium chloride must be CaCI. So, this answer is wrong

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The A-36 steel pipe has a 6061-T6 aluminum core. It is subjected to a tensile force of 200 kN. Determine the average normal stre

rodikova [14]

Answer:

In the steel: 815 kPa

In the aluminum: 270 kPa

Explanation:

The steel pipe will have a section of:

A1 = π/4 * (D^2 - d^2)

A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2

The aluminum core:

A2 = π/4 * d^2

A2 = π/4 * 0.7^2 = 0.3848 m^2

The parts will have a certain stiffness:

k = E * A/l

We don't know their length, so we can consider this as stiffness per unit of length

k = E * A

For the steel pipe:

E = 210 GPa (for steel)

k1 = 210*10^9 * 0.1178 = 2.47*10^10 N

For the aluminum:

E = 70 GPa

k2 = 70*10^9 * 0.3848 = 2.69*10^10 N

Hooke's law,

Δd = f / k

Since we are using stiffness per unit of length we use stretching per unit of length,

ε = f / k

When the force is distributed between both materials will stretch the same length,

f = f1 + f2

f1 / k1 = f2/ k2

Replacing:

f1 = f - f2

(f - f2) / k1 = f2 / k2

f/k1 - f2/k1 = f2/k2

f/k1 = f2 * (1/k2 + 1/k1)

f2 = (f/k1) / (1/k2 + 1/k1)

f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN

f1 = 200 - 104 = 96 kN

Then we calculate the stresses:

σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa

σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa

5 0

3 years ago

The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?

quester [9]

Answer:

$Ksp=1.07x10^{-8}$

Explanation:

Hello,

In this case, the dissociation reaction is:

$PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)$

For which the equilibrium expression is:

$Ksp=[Pb^{2+}][I^-]^2$

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

$Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M$

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

$[Pb^{2+}]=1.39x10^{-3}M$

$[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M$

Thereby, the solubility product results:

$Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}$

Regards.

6 0

4 years ago

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BlackZzzverrR [31]

Answer:

the independant variable

Explanation:

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3 years ago

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Alinara [238K]

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3 years ago

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insens350 [35]

Answer:

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Explanation:

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3 years ago