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3 years ago

How many atoms of group 17 element would be needed react with one atom of a group 2 element? Explain

2 answers:

5 0

2 atoms of group 7 elements, because group 2 elements have 2 valence electrons and group 7 need just one atoms to stabilize.

shusha [124]3 years ago

3 0

Only one atom because atom with 17 electrons need that one electron to become stable

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What substances pass through a leaf's stomata?

Lera25 [3.4K]

Oxygen and carbon dioxide

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3 years ago

Read 2 more answers

Which substance is not a structural isomer of hexyne?

stich3 [128]

Answer:

c) 3,3-dimethylpent-1-yne

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2 years ago

The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its

Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

$t_{\frac{1}{2}} =$ $\frac{ln 2}{k}$

$t_{\frac{1}{2}} = \frac{0.693}{k}$

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

$Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}$

$k = \frac{0.693}{5230 years}$

$k = 1.325 * 10^{-4} yr^{-1}$

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

$[A] = [A_{0}]e^{-kt}$

$\frac{[A]}{[A_{0}]} = e^{-kt}$

$\frac{35}{100} = e^{-(1.325 *10^{-4})t}$

$ln(0.35) = -(1.325 *10^{-4})(t)$

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0

3 years ago

Use the reaction given below to solve the problem that follows: Calculate the mass in grams of aluminum oxide produced by the re

bearhunter [10]

Answer: 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles$

The balanced chemical equuation is:

$4Al+3O_2\rightarrow 2Al_2O_3$

According to stoichiometry :

4 moles of $Al$ produce == 2 moles of $Al_2O_3$

Thus 0.556 moles of $Al$ will produce= $\frac{2}{4}\times 0.556=0.278moles$ of $Al_2O_3$

Mass of $Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g$

Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.

7 0

2 years ago

Plz help me solve this question is it A,B,C or D

Rashid [163]

Answer:

B - To increase the rate of the reaction

Explanation:

Catalysts speed up the reaction without being reactants or products, so aren't used up in the reaction.

6 0

2 years ago