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3 years ago

(Please answer as quickly as possible for brainliest, thanks :))

Chemistry

2 answers:

Eddi Din [679]3 years ago

4 0

A. actinides do not occur in nature.

damaskus [11]3 years ago

3 0

Hello,

The answer is option A <span>Actinides.

Reason:

The answer is </span><span>Actinides because they are human made. Its not option B because some of those are minerals. Its not C because gases are caused by nature. Its also not D because nonmetals are also made by nature.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>

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3 years ago

El aguacate al oxidarse, ¿que fenómeno es?, ¿físico, químico o alotropicos?

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3 years ago

Water is heated to boiling and water vapor (steam) is released. Chemical change or physical change?

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3 years ago

When a solution containing 1.4000 g of Ba(NO3)2 and 2.4000 g of HSO3NH2 is boiled, a precipitate forms. One possible identity fo

Georgia [21]

Answer:

See explanation for detailed solution

Explanation:

The balanced reaction equation is Ba(NO3)2 + 2HSO3NH2 → Ba(SO3NH2)2 + 2HNO3

Number of moles of Ba(NO3)2 = 1.4 g/ 261.337 g/mol = 5.36 × 10^-3 moles

From the reaction equation;

1 mole of Ba(NO3)2 yields 1 mole of Ba(SO3NH2)2

5.36 × 10^-3 moles of Ba(NO3)2 yields 5.36 × 10^-3 moles of Ba(SO3NH2)2

For HSO3NH2

Number of moles = 2.4g/97.10 g/mol =0.0247 moles

2 moles of HSO3NH2 yields 1 mole of Ba(SO3NH2)2

0.0247 moles of HSO3NH2 yields 0.0247 ×1/2 = 0.0137 moles

Hence, Ba(NO3)2 is the limiting reactant

The theoretical yield of Ba(SO3NH2)2 is 5.36 × 10^-3 moles × 329.4986 g/mol = 1.766 g

Number of moles = mass/ molar mass

Molar mass = mass/ number of moles

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4 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

3 0

3 years ago