Tin+Chlorine -> Chlorine -> Tin(IV) chloride

How many moles of H2 would be required to produce 37.0 moles of water?

svp [43]

Hii there !

The reaction accompanying production of water in this case would be -

$\green{ \underline { \boxed{ \sf{H_2+ \frac{1}{2}O_2\longrightarrow H_2O}}}}$

Thus ,by observing the reaction, we can conclude that 1 mole of Hydrogen (H $_2$ )react with 1/2 mole of oxygen(O $_2$ )to produce 1 mole of water (H $_2$ O)

$\implies$ Therefore, 37.0 moles of hydrogen(H $_2$ ) will be required to produce 37.0 moles of water.

Hope it helps.

7 0

2 years ago

Show how water can serve as a base by showing the chemical equation of water reacting with hydrochloric acid. g

goldfiish [28.3K]

So, water reacts with hydrochloric acid in the following formula

H2O + HCl —-> H3O+ + Cl-

We can visualize that when the two react, the hydrogen ions is taken on by the water molecule. This satisfies one of the definitions for a base

Bronsted acids = anything that donates a proton (H+ ion)

Bronsted bases = anything that accepts a proton (H+ ion)

So, as we can see, that is exactly what is happening. The Cl- and H+ detach and then the water takes on that extra H+.

H3O+ is what we call a hydronium ion

8 0

3 years ago

What is the function of hemoglobin in the body?

aleksley [76]

Main function of haemoglobin in the body is to transport oxygen to every cell/organ of the body

Hope this helps!!

7 0

4 years ago

Read 2 more answers

At a given temperature, 4.06 atm of H2 and 3.5 atm of Cl2 are mixed and allowed to come to equilibrium. The equilibrium pressure

Llana [10]

<u>Answer:</u> The value of $K_p$ for the given chemical reaction is 0.1415

<u>Explanation:</u>

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as $K_p$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

The expression for $K_p$ is written as:

$K_p=\frac{p_{C}^cp_{D}^d}{p_{A}^ap_{B}^b}$

For the given chemical equation:

$H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g)$

The expression for $K_p$ for the following equation is:

$K_p=\frac{(p_{HCl})^2}{(p_{H_2)}(p_{Cl_2})}$

We are given:

$p_{HCl}=1.418atm\\p_{H_2}=4.06atm\\p_{Cl_2}=3.5atm$

Putting values in above equation, we get:

$K_p=\frac{(1.418)^2}{(4.06)\times (3.5)}\\\\K_p=0.1415$

The value of $K_p$ for the given chemical reaction is 0.1415

5 0

3 years ago

he standard enthalpies of formation for S (g), F (g), SF4 (g), and SF6 (g) are +278.8, +79.0, -775, and -1209 kJ per mole, respe

JulsSmile [24]

Answer:

+60.54 kJ/mol

Explanation:

The enthalpy of formation of a compound is the sum of the enthalpies of the formation of its constituents and the bond of them. To form a compound, energy must be lost, so, they're negative.

For SF4, the enthalpy is formed by the energy of one S, two F, and 4 S-F bond (Hb):

H = - (278.8 + 4*79.0 + 4*Hb)

-775 = -(594.8 + 4Hb)

594.8 + 4Hb = 775

4Hb = 180.2

Hb = +45.05 kJ/mol

For SF6, the enthalpy is formed by the energy of one S, six F, and 6 S--F bonds (Hb):

H = -(278.8 + 6*79.0 + 6*Hb)

-1209 = -(752.8 + 6Hb)

752.8 + 6Hb = 1209

6Hb = 456.2

Hb = +76.03 kJ/mol

Thus, the energy of S--F bond must be the average of these two:

(45.05 + 76.03)/2 = +60.54 kJ/mol

3 0

3 years ago