K+ and Po43- formula

20 ml of a 1.0 m hcl solution are added to 3.0 l of acid solution with a ph of 5.35. the ph of the mixture rises to 5.33 was the

SashulF [63]

You may suppose you have a 0.1 M solution of NH3, from:

NH4Cl + NaOH > NH3 + H2O.

Then you can compute the pH from the concentration of NH3 and its pKb.

The concentration is high enough to use the simplified formula:

[OH] = sqr(Kb*conc)

3 0

3 years ago

PLEASE HELP ME!!!

Kazeer [188]

722391465.060241 MHz

sbbshahshaja

3 0

3 years ago

You are running a rather large scale reaction where you prepare the grignard reagent phenylmagnesium bromide by reacting 210.14

almond37 [142]

Answer:

We would expect to form 7.35 moles of grignard reagent.

Explanation:

Step 1: Data given

Mass of magnesium = 210.14 grams

Volume bromobenzene = 772 mL

Density of bromobenzene = 1.495 g/mL

Molar mass of Mg = 24.3 g/mol

Molar mass of bromobenzene = 157.01 g/mol

Step 2: The balanced equation

C6H5Br + Mg ⇒ C6H5MgBr

Step 3: Calculate mass of bromobenzene

Mass bromobenzene = density bromobenzene * volume

Mass bromobenzene = 1.495 g/mL * 772 mL

Mass bromobenzene = 1154.14 grams

Step 4: Calculate number of moles bromobenzene

Moles bromobenzene = mass bromobenzene / molar mass bromobenzene

Moles bromobenzene = 1154.14g / 157.01 g/mol

Moles bromobenzene = 7.35 moles

Step 5: Calculate moles of Mg

Moles Mg = 210.14 grams /24.3 g/mol

Moles Mg = 8.65 moles

Step 6: The limiting reactant

The mole ratio is 1:1 So the bromobenzene has the smallest amount of moles, so it's the limiting reactant. It will be completely consumed ( 7.35 moles). Magnesium is in excess, There will react 7.35 moles. There will remain 8.65 - 7.35 = 1.30 moles

Step 7: Calculate moles of phenylmagnesium bromide

For 1 mole of bromobenzene, we need 1 mole of Mg to produce 1 mole of phenylmagnesium bromide

For 7.35 moles bromobenzene, we have 7.35 moles phenylmagnesium bromide

We would expect to form 7.35 moles of grignard reagent.

5 0

3 years ago

How do magnetic forces repel or attract? make it short and simple 25 points

stiks02 [169]

Answer:

When you place the north pole of one magnet near the south pole of another magnet, they are attracted to one another.

Explanation:

6 0

3 years ago

Read 2 more answers

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

= (-3399.6) + (1560)

= -1839.6 KJ

5 0

3 years ago