Why can color alone be used to identify most minerals?

Answers

Answer 1

Answer: Because a lot of minerals have the same color

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Samarium-146 has a half-life of 103.5 million years. After 1.035 billion years, how much samarium-146 will remain from a 205-g sample?0.200 g
0.400 g
20.5 g
103 g

Answers

Answer : The correct option is, 0.200 g

Solution :

As we know that the radioactive decays follow first order kinetics.

First we have to calculate the rate constant of a samarium-146.

Formula used :

$t_(1/2)=(0.693)/(k)$

Putting value of $t_(1/2)$ in this formula, we get the rate constant.

$103.5* 10^6=(0.693)/(k)$

$k=6.6* 10^(-9)year^(-1)$

Now we have to calculate the original amount of samarium-146.

The expression for rate law for first order kinetics is given by :

$k=(2.303)/(t)\log(a)/(a-x)$

where,

k = rate constant = $6.6* 10^(-9)year^(-1)$

t = time taken for decay process = $1.035* 10^(9)years$

a = initial amount of the samarium-146 = 205 g

a - x = amount left after decay process = ?

Putting values in above equation, we get the value of initial amount of samarium-146.

$6.6* 10^(-9)=(2.303)/(1.035* 10^(9))\log(205)/(a-x)$

$a-x=0.200g$

Therefore, the amount left of the samarium-146 is, 0.200 g

Ans: 0.200 g

Given:

Half life of Sm-146 = t1/2 = 103.5 million years

Time period, t = 1.035 billion years = 1035 million years

Original mass of sample, [A]₀ = 205 g

To determine:

Amount of sample after t = 1035 million years

Explanation:

The rate of radio active decay is given as:

$A(t) = A(0)e^(-0.693t/t1/2) \n\n= 205 g * e^{(0.693*1035)/(103.5) } \n\n= 0.200 g$

Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half-lives of polonium-210 occur in 966 days? How much polonium is in the sample 966 days later?

Answers

The answer is a) 7 half-lives and b) 0.78% (or as fraction 1/128) of the starting amount.

a) If half-life of Polonium-210 is 138 days, to calculate how many half-lives occur in 966 days, we will simply divide them: 966/138 = 7
So, 7 half-lives occur in 966 days.

b) To calculate the remaining amount, we will use the formula:
$(1/2)^(n) =x$
where n is the number of half-lives, and x is the remaining amount in decimals, and (1/2) is half-life.

We've already found that n = 7, so replace it in the formula:
$(1/2)^(7) =0.0078 =$ 0.78% = 1/128

Answer : The amount of polonium in the sample 966 days later is, 0.24 g

Solution :

Polonium-210 is a radioactive element.

Formula used :

$N(t)=N_o* ((1)/(2))^{(t)/(t_(1/2))$

where,

$N(t)$ = the amount of polonium-210 remaining after 't' days

$N_o$ = the initial amount of polonium-210 = 31 g

t = time = 966 days

$t_(1/2)$ = half-life of the polonium-210 = 138 days

Now put all the given values in the above formula, we get

$N(t)=(31g)* ((1)/(2))^{(966)/(138)}$

$N(t)=0.24g$

Therefore, the amount of polonium in the sample 966 days later is, 0.24 g

How many moles of nitrogen trifluoride (NF3) can be produced from 9.65 mole of Fluorine gas (F2)N2 + 3F2 -----------> 2NF3

Answers

Answer:

6.43 moles of NF₃.

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3F₂ —> 2NF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.

Thus, 6.43 moles of NF₃ were obtained from the reaction.

Which represents the copernican model that is the most similar to that of Aristarchus?

Answers

Answer: A

Explanation: A is the answer

Final answer:

The Copernican model that is most similar to Aristarchus' . It is a simple and accurate heliocentric model that explains planetary motion with a small set of rules and a single underlying force.

Explanation:

The Copernican model that is most similar to that of Aristarchus is represented by Figure 6.31(b). In this model, Earth and other planets revolve around the Sun. It is a simpler and more accurate heliocentric model, similar to Aristarchus' idea of a heliocentric solar system. The Copernican model explains planetary motion with a small set of rules and a single underlying force, demonstrating the breadth and simplicity of the laws of physics.

Learn more about Copernican model similar to Aristarchus here:

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What is the volume of 70.0 g of ether if density os 0.70 g/mL

Answers

$D = m / V$

$0.70 g/mL = (70.0g)/(V)$

$V = (m)/(D)$

$V = (70.0g)/(0.70g/mL)$

$V = 100 mL$

hope this helps!

Select the correct answer. Which statement describes the acid found in vinegar (acetic acid)? A. It tastes bitter. B. It feels slippery. C. It changes red litmus paper to blue. D. It reacts with Mg to produce H2. E. It releases OH- in a solution.

Answers

Answer: Option (D) is the correct answer.

Explanation:

The acetic acid, $(CH_(3)COOH)$ present in vinegar reacts wit magnesium and therefore, it releases hydrogen gas and magnesium acetate.

The chemical reaction will be as follows.

$2CH_(3)COOH + 2Mg \rightarrow 2Mg(CH_(3)COO) + H_(2)$

An acid is a substance which has pH less than 7 and changes blue litmus red. Acids taste sour as they are acidic in nature and releases hydrogen ions.

Whereas bases have pH greater than 7 and changes red litmus into blue.Bases taste bitter and releases hydroxide ions.

Thus, we can conclude that the statement it reacts with Mg to produce $H_(2)$ best describes the acid found in vinegar (acetic acid).

Answer:

D. It reacts with Mg to produce H2.

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Which molecule, when added to a solution, would give the solution the highest pH?HCl H2SO4 KOH H2O

Which statement best explains why heating a liquid affects its viscosity?

Air is made up of different gases, such as oxygen, nitrogen, and carbon dioxide. Which statement best describes these three components of air? A)They are all classified as pure substances. B)They cannot react with another substance. C)They are chemically bonded to one another. D)They can be classified as elements.