Which three terms are needed to describe the energy a base jumper has as she falls toward the ground ? APEX

Balancing the reaction by oxidation number method k2cr2o7+sncl2+hcl

Mumz [18]

Answer:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

Explanation:

The products of this reaction are given by:

$K_2Cr_2O_7 (aq) + SnCl_2 (aq) + HCl (aq)\rightarrow KCl (aq) + SnCl_4 (aq) + CrCl_3 (aq) + H_2O (l)$

Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

$Cr_2O_7^{2-} (aq)\rightarrow Cr^{3+} (aq)$

The following steps should be taken:

balance the main element, chromium: multiply the right side by 2 to get 2 chromium species on both side:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)$

balance oxygen atoms by adding 7 water molecules on the right:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the hydrogen atoms by adding 14 protons on the left:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

Similarly, tin(II) cation becomes tin(IV) cation:

$Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-$

Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

$3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-$

Add them together:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 3 Sn^{2+} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l) + 3 Sn^{4+} (aq)$

Adding the ions spectators:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

7 0

3 years ago

Which type of heat exchanger shell and tube or plate heat exchanger is suitable for evaporation?

klasskru [66]

Answer:

shell and tube type heat exchanger

Explanation:

for evaporation the shell and tube type heat exchanger is best suited.

in the plate heat exchanger there is gaskets in between every part so this part become weak part in heat echanger and there is possibilities of leakage through this part, there is no such problem in shell and tube type.
the plate type cant be used when there is high temperature and high pressure drop but shell and tube type can be used
in evaporation there the liquids change into vapors due to which there is sudden change in pressure and in which plate type is not used because there is chances of leakage

5 0

2 years ago

Tellurium is a silvery-white, brittle element. Its electrical conductivity varies, but I’m creases slightly with exposure to sun

Lina20 [59]

B) Tellurium is a metalloid. Metalloids in some classifications are also or alternatively called “semi-metals.”

5 0

3 years ago

Read 2 more answers

The first three steps in the formation of the sun are listed below.

taurus [48]

C is the answer, I think

4 0