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Sedaia [141]
3 years ago
9

A 1.00 liter solution contains 0.25 M ammonia and 0.32 M ammonium bromide. If 0.080 moles of barium hydroxide are added to this

system, indicate whether the following statements are true or false. (Assume that the volume does not change upon the addition of barium hydroxide.) _______ A. The number of moles of NH3 will increase. _______ B. The number of moles of NH4 will decrease. _______ C. The equilibrium concentration of H3O will remain the same. _______ D. The pH will decrease. _______ E. The ratio of [NH3] / [NH4 ] will increase.
Chemistry
1 answer:
puteri [66]3 years ago
8 0
I need points i’m sorry ahhh
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Write the chemical formula of the following:
Elan Coil [88]

Answer:

1.Ca(OH)2

2.Mg(OH)2

3.Al2SO4

4.Na2CO3

5.Zn3(PO4)2

6.LiNO3

7.CuCO3

8. K2CO3

9.MgSO4

10.Ca3(PO4)2

5 0
3 years ago
A 25.225 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 75.815 g of wa
Rudik [331]

Answer:

1.43 (w/w %)

Explanation:

HCl reacts with NH3 as follows:

HCl + NH3 → NH4+ + Cl-

<em>1 mole of HCl reacts per mole of ammonia.</em>

Mass of NH3 is obtained as follows:

<em>Moles HCl:</em>

0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>

<em>Mass NH3 in the aliquot:</em>

3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.

Mass of sample + water = 22.225g + 75.815g = 98.04g

Dilution factor: 98.04g / 14.842g = 6.6056

That means mass of NH3 in the sample is:

0.0545g * 6.6056 = 0.36g NH3

Weight percent is:

0.36g NH3 / 25.225g * 100

<h3>1.43 (w/w %)</h3>
6 0
3 years ago
Calculate the mass in grams of 202 atoms of iron
Sergeeva-Olga [200]

Answer:

To calculate the number of atoms in a sample, divide its weight in grams by the amu atomic mass from the periodic table, then multiply the result by Avogadro's number: 6.02 x 10^23. Set up Equation Express the relationship of the three pieces of information you need to calculate the number of atoms in the sample in the form of an equation.

4 0
3 years ago
Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r
Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

7 0
3 years ago
If the density of an object is 5.2g/cm3, and it’s volume is 3.7 cm3, what is it’s mass?
lord [1]
Here's the equation you use: Density = mass/volume 

1) 5.2g/cm^3 = m/3.7cm^3 

2) m = 5.2g/cm^3 x 3.7cm^3 

3) m = 19.24g 

You can check the answer by plugging it in 

19.24g/3.7cm^3 
= 5.2g/cm^3
5 0
3 years ago
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