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3 years ago

13

Which statement is true of the carbon atoms that make up a diamond

1 answer:

Lunna [17]3 years ago

6 0

Crystal<span> of tetrahedrally bonded carbon atoms in a </span>covalent network lattice<span> witch </span>crystallizes<span> into the diamond </span>lattice.

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Given the following equation, how many grams of PbCO3 will dissolve when exactly 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3?

9966 [12]

Calculating for the moles of H+
1.0 L x (1.00 mole / 1 L ) = 1 mole H+

From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3

Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
0.5 moles PbCO3 x (267 g PbCO3 / 1 mole PbCO3) = 84.5 g PbCO3

4 0

3 years ago

Read 2 more answers

What is the pH of 0.000134 M solution of HCI?

mafiozo [28]

Answer: The pH will be 3.87

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$HCl\rightarrow H^++Cl^{-}$

According to stoichiometry,

1 mole of $HCl$ gives 1 mole of $H^+$

Thus $0.000134$ moles of $HCl$ gives = $\frac{1}{1}\times 0.000134=0.0001342$ moles of $H^+$

Putting in the values:

$pH=-\log[0.000134]$

$pH=3.87$

Thus the pH will be 3.87

5 0

3 years ago

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Three cards with holes are arranged in a straight line. A light is shined through the first card’s hole and travels through all

igomit [66]

C) that light travels in a straight line.

8 0

4 years ago

Read 2 more answers

Round 1532.2364 to 3 significant figures.

mote1985 [20]

Answer:

It would be 1530

7 0

3 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