Caesium (55Cs) has 40 known isotopes, making it, along with barium and mercury, one of the elements with the most isotopes. The atomic masses of these isotopes range from 112 to 151. Only one isotope, 133Cs, is stable. The longest-lived radioisotopes are 135Cs with a half-life of 2.3 million years, ... It constitutes most of the radioactivity still left from the Chernobyl accident ...

3 0

2 years ago

What is the molarity of a solution of sulfuric acid if 49.0 mL of it are neutralized by a titration

vaieri [72.5K]

Answer:

THE MOLARITY OF THE ACID IS 0.232M

Explanation:

Molarity is the number of moles of solute per liter of a solution.

In titration, the molarity of the acid and the molarity of the base are related by this equation:

Ca VA / CbVb = na / nb

Ca = concentration of the acid = ?

Va = volume of the acid = 49 mL

Cb = concentration of the base = 0.333M

Vb = volume of the base = 68.4mL

Na = number of mole of acid = 1

Nb = number of mole of base = 2

Equation for the reaction:

H2SO4 + 2NaOH --------> Na2SO4 + 2H2O

Solving for the molarity of the acid, we have :

Ca = CbVbNa / VaNb

Ca = 0.333M * 68.4mL * 1 / 49mL *2

Ca = 22.7772 * 10^-3 / 98 *10^-3

Ca = 0.232M

The molarity of the acid is therefore 0.232M.

8 0

3 years ago

Read 2 more answers

What is the concentration in molarity of S2O32- (aq) in a solution prepared by mixing 150 mLmL of 0.149 MM Na2S2O3 (aq) with eno

Artist 52 [7]

Answer:

0.0890 M

Explanation:

Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:

C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL

C2 = 0.149 x 150/250

= 0.089 M

To determine the concentration of S2O32- (aq), consider the equation:

$Na_2S_2O_3 => 2Na^+_{(aq)} + S_2O^2^-_3_{(aq)}$

The concentration of Na2S2O3 and S2O32- (aq) is 1:1

Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.

To 3 significant figures = 0.0890 M

7 0

3 years ago

Consider the reaction 4PH3(g) → P4(g) + 6H2(g) At a particular point during the reaction, molecular hydrogen is being formed at

Svetllana [295]

Answer:

The rate at which $P_4$ is being produced is 0.0228 M/s.

The rate at which $PH_3$ is being consumed is 0.0912 M/s.

Explanation:

$4PH_3\rightarrow P_4(g)+6H_2(g)$

Rate of the reaction : R

$R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}$

The rate at which hydrogen is being formed = $\frac{d[H_2]}{dt}=0.137 M/s$

$R=\frac{1}{6}\frac{d[H_2]}{dt}$

$R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s$

The rate at which $P_4$ is being produced:

$R=\frac{1}{1}\frac{d[P_4]}{dt}$

$0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}$

The rate at which $PH_3$ is being consumed :

$R=\frac{-1}{4}\frac{d[PH_3]}{dt}$

$0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}$

$\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s$

6 0

3 years ago