Convert 3.09 x 10~3 m = nm

Simplify the numercial expression 13-2x5

Tema [17]

Answer:

3

Explanation:

2x5 is 10 then 13-10 is 3

4 0

3 years ago

Read 2 more answers

Balance the following redox equation using the smallest integers possible and select the correct coefficient for the hydrogen su

goldfiish [28.3K]

Answer:

Coefficient of $HSO_{3}^{-}$ is 5

Explanation:

Oxidation: $MnO_{4}^{-}(aq)\rightarrow Mn^{2+}(aq)$

Balance O and H in acidic medium: $MnO_{4}^{-}(aq)+8H^{+}(aq)\rightarrow Mn^{2+}(aq)+4H_{2}O(l)$
Balance charge : $MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow Mn^{2+}(aq)+4H_{2}O(l)$ .............(1)

Reduction: $HSO_{3}^{-}\rightarrow SO_{4}^{2-}$

Balance O and H in acidic medium : $HSO_{3}^{-}(aq)+H_{2}O(l)\rightarrow SO_{4}^{2-}(aq)+3H^{+}(aq)$
Balance charge : $HSO_{3}^{-}(aq)+H_{2}O(l)-2e^{-}\rightarrow SO_{4}^{2-}(aq)+3H^{+}(aq)$ ..............(2)

$[2\times eq(1)]+[5\times eq(2)]$ -

Balanced equation: $2MnO_{4}^{-}(aq)+5HSO_{3}^{-}(aq)+H^{+}(aq)\rightarrow 2Mn^{2+}(aq)+5SO_{4}^{2-}(aq)+3H_{2}O(l)$

Coefficient of $HSO_{3}^{-}$ is 5

8 0

3 years ago

The red outlined elements have what major characteristics in common?

CaHeK987 [17]

One major characteristic they have in common is they are metals.

5 0

3 years ago

What will the concentration of PCl5 be when equilibrium is reestablished after addition of 1.31 g Cl2? PCl5(g) ⇆ PCl3(g) + Cl2(g

Masteriza [31]

Answer:

The new concentration of PCl5 will be 0.01953 M

Explanation:

Step 1: Data given

Mass of Cl2 added = 1.31 grams

Molar mass Cl2 = 70.9 g/mol

Original Equilibrium Mixture:

3.42 g PCl5

4.86 g PCl3

3.59 g Cl2

Volume = 1.0 L

Step 2: The balanced equation

PCl5(g) ⇆ PCl3(g) + Cl2(g)

Step 3: Calculate the original moles and molarity

Moles = mass / molar mass

Moles PCL5 = 3.42 grams / 208.24 g/mol

Moles PCl5 = 0.0164 moles

[PCl5] = 0.0164 M

moles PCl3 = 4.86 grams / 137.33 g/mol

moles PCl3 = 0.0354 moles

[PCl3] = 0.0354 M

moles Cl2 = 3.59 grams / 70.9 g/mol

moles Cl2 = 0.0506 moles

[Cl2] = 0.0506 M

the new mass Cl2 = 3.59 + 1.31 = 4.9 grams

moles Cl2 = 0.0691 moles

[Cl2]= 0.0691 M

The new concentration at the equilibrium

[PCl5] = 0.0164 + X M

[PCl3 ] = 0.0354 - X M

[Cl2] = 0.0691 - X M

Step 4: Calculate Kc

Kc = [Cl2][PCl3] / [PCl5]

Kc = (0.0506*0.0354)/0.0164

Kc = 0.109

Step 5: Calculate [PCl5]

Kc = 0.109 = ((0.0691 - X)(0.0354 - X)) / (0.0164 + X)

X = 0.00313

[PCl5] = 0.0164 + 0.00313 M = 0.01953 M

[PCl3 ] = 0.0354 - 0.00313 M = 0.03227 M

[Cl2] = 0.0691 - 0.00313 M = 0.06597

The new concentration of PCl5 will be 0.01953 M

6 0

3 years ago

If 57.0 g of b2o3 is added to 44.7 g of cl2 and 68.8 g of c, what is the theoretical yield of boron trichloride?

sdas [7]

(57.0 g B2O3 / (69.6202 g B2O3/mol) x (4mol BCI3 / 2 mol B2O3) = 1.64 mol BC13

(44.7 g C12) / (70.9064 g C12/mol) x (4mol BCI3 / 6mol C12) = 0.42027 mol BC13

(68.8 g C) / (12.01078 G C/mol) x (4mol BCI3 / 3 mol C) = 7.63 mol BCI3

C12 is the limiting reactant.

(0.42027 mol BCI3) X (117 . 170 g BCI3/mol) = 49.2 g BCI3 in theory.

6 0

3 years ago