John was considered to be a pillar in the early church?

disa [49]

Answer:

6H2 + P4→ 4PH3

Explanation:

Phosphorus has 4 in it and hydrogen has 3 in it. in order to balance it, we have to put 4 in front of phosphine so that the phosphorus on the product side has an equal amount as to the one on the reactant side.

the only one left to balance is hydrogen and so in order to balance it we put a 6 on h2 because the hydrogen in the product size becomes 12 (4 * 3).

therefore the hydrogen on the reactant side becomes 12 as well (6 * 2)

8 0

3 years ago

What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?

Ymorist [56]

Answer:

$\% diss = 50\%$

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

$HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}$

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

$Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}$

Thus, it is possible to find x given the pH as shown below:

$x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M$

So that we can calculate the initial concentration of the acid:

$\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\$

$[HA]_0=3.64x10^{-5}M$

Therefore, the percent dissociation turns out to be:

$\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%$

Best regards!

6 0

2 years ago

Drag the tiles to the correct boxes. Not all tiles will be used.

gizmo_the_mogwai [7]

Answer:

metre measures length

kilogram measures mass

time is seconds

Explanation:

I have matched correctly. Just check SI units and there measures.

4 0

2 years ago

Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride (TiCl4

vova2212 [387]

Answer:

a) 226.6 grams of Cl₂

b) 19.2 grams of C

c) 303.2 grams of TiCl₄ and 70.4 grams of CO₂

Explanation:

The balanced chemical reaction is the following:

TiO₂(s) + C(s) + 2 Cl₂(g) → TiCl₄(s) + CO₂(g)

(a) What mass of Cl₂ gas is needed to react with 1.60 mol TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 2 moles of Cl₂. So, the stoichiometric ratio is 2 mol Cl₂/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of Cl₂ we need:

1.60 mol TiO₂ x 2 mol Cl₂/1 mol TiO₂ = 3.2 mol Cl₂

Now, we convert from moles to mass by using the molecular weight (MW) of Cl₂:

MW(Cl₂) = 35.4 g/mol x 2 = 70.8 g/mol

mass of Cl₂= 3.2 mol x 70.8 g/mol = 226.6 g

Therefore, 226.6 grams of Cl₂ are needed to react with 1.6 mol of TiO₂.

(b) What mass of C is needed to react with 1.60 mol of TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 1 moles of C(s). So, the stoichiometric ratio is 1 mol C/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of C(s) we need:

1.60 mol TiO₂ x 1 mol C(s)/1 mol TiO₂ = 1.60 mol C(s)

So, we convert the moles of C(s) to grams as follows:

MW(C) = 12 g/mol

1.60 mol x 12 g/mol = 19.2 g C(s)

Therefore, a mass of 19.2 grams of C is needed to react with 1.60 mol of TiO₂.

(c) What is the mass of all the products formed by reaction with 1.60 mol of TiO₂?

From the chemical equation, we can notice that 1 mol of TiO₂ produces 1 mol of TiCl₄ and 1 mol of CO₂. So, from 1.60 moles of TiO₂, 1 mol of each product will be produced:

1 mol TiO₂/1 mol TiCl₄ ⇒ 1.60 mol TiO₂/1.60 mol TiCl₄

1 mol TiO₂/1 mol CO₂ ⇒ 1.60 mol TiO₂/1.60 mol CO₂

Finally, we convert the moles to grams by using the molecular weight of each compound:

MW(TiCl₄) = 47.9 g/mol Ti + (35.4 g/mol x 4 Cl) = 189.5 g/mol

1.60 mol x 189.5 g/mol = 303.2 g

MW(CO₂) = 12 g/mol C + (16 g/mol x 2 O) = 44 g/mol

1.60 mol x 44 g/mol = 70.4 g

Therefore, from the reaction of 1.60 mol of TiO₂ are formed 303.2 grams of TiCl₄ and 70.4 grams of CO₂.

3 0

3 years ago

A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following

marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}$

V₂ = 4,00L

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}$

V₂ = 16,00L

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

$\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}$

V₂ = 12,31L

I hope it helps!

6 0

3 years ago