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3 years ago

The ph of a solution for which [OH]= 1.0 x 10^-4 is? A 10 B 14 C 11 D 9 E 7

1 answer:

8 0

To find pH, use the following formula ---> pH= - log [H+]

so first we need to calculate the [H+] concentration using the OH concentration. to do this, we need to use this formula--> 1.0x10-14= [H+] X [OH-], so we solve for H+ and plug in

[H+]= 1.0X10-14/[OH-]---> 1.0 x 10-14/ 1.0 x 10-4= 1.0 x 10-10

now that we have the H+ concentration, we can solve of pH

pH= -log (1.0x10-10)= 10

answer is A

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One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

How can you tell that a gas is a halogen? ___?

OLEGan [10]

If it is located at the second to last row of the periodic table (the halogen family), has seven electrons on it's outer shell, and has an oxidation number of -1, it is a halogen.
Hope this helps : D

8 0

3 years ago

Alex787 [66]

Answer:

The reaction of an acid and a base is called a neutralization reaction. ... However, in the reaction between HCl(aq) and Mg(OH) 2(aq), additional molecules of HCl and ... First, we will write the chemical equation with the formulas of the reactants ... The chloride ions are the only spectator ions I THINKS

Explanation:

4 0

2 years ago

The concentration of a trace mineral is found to be 2.53 PPB in an aqueous solution. what volume of solution contains 2.53 g of

FinnZ [79.3K]

Answer:

4. A concentration of a trace mineral was found to be 2.53 ppb in an aqueous solution. A. What mass of solution contains 2.53 g of the mineral? (1 point - Knowledge, 2 points - Inquiry, 1 point - Communication) m=? C = 2.53 ppm or 2.53X106 n = Con ܗ B. What volume of solution contains 2.53 g of the mineral? (1 point - Knowledge, 2 points - Inquiry, 1 point - Communication) VE? C: 2,53 ppm ns Cen

7 0

1 year ago

¿Cuál de las siguientes configuraciones globales corresponde a un elemento químico que se comporta como metal? 1 punto [He]2s2 2

yuradex [85]

Answer:

[Ne]3s2

Explanation:

ahora tenemos que mirar cada una de las configuraciones electrónicas de cada átomo de cerca antes de tomar una decisión.

considerando la configuración electrónica más externa de cada una de las especies mostradas;

para la primera configuración, ns2 np6 corresponde a un gas noble.

para la segunda configuración ns2 np3 corresponde a un elemento no metálico del grupo 5.

para la tercera configuración, ns2 corresponde a un elemento metálico del grupo 2.

para la cuarta configuración, ns2 np4 corresponde a un elemento no metálico del grupo 6

5 0

3 years ago