(NH4)3P is it Ionic or Molecular??​

The amount of diamine silver chloride = 8.87 g

Given

10 g AgCl

1.00 L of 0.100 M NH3

Required

the amount of diamine silver

Reaction

AgCl + 2 NH₃ → [Ag(NH₃)₂]Cl

mol AgCl :

= mass : MW

= 10 g : 143,32 g/mol

= 0.0698

mol NH₃ :

= M x V

= 0.1 x 1

= 0.1

NH₃ as a limiting reactant

mol [Ag(NH₃)₂]Cl based on NH₃ :

= 1/2 x mol NH₃

= 1/2 x 0.1

= 0.05

Mass diamine silver :

= 0.05 x  177.3822 g/mol

= 8.87 g

Answer:

SrO = Sr + O - Chemical Equation Balancer.

Explanation:

Answer:

Matter is made up of particles whose properties determine the observable characteristics of matter and its reactivity.

Explanation:

Answer:

2 curies

Explanation:

This is actually pretty easy to explain.

The Radium-226 which is an isotope of the radium, have an activity of curie. This is basically used in radioactivity of certain elements.

As in both cases we have Radium -226 and we are only varying the mass of the Radium, then, we can just do a rule of 3 to calculate this:

1 g Ra-226 --------> 1 Curie

2 g Ra-226 -------->  X Curie

Solving for X:

X = 2 g * 1 Curie / 1 g

You can also see this by taking 2 grams of Ra per separate. On one side you have 1 g of Ra-226 and on the other side, you have another 1 g of Ra-226. Both have 1 Cury of activity. If you put both on the same side, you will have 1 Cury of 1 isotope plus 1 cury of the other isotope, final result? 2 curies.

Hope this helps

Answer:

C. Ionic-Li and S are two nonmetals

Answer:

antimony

Explanation:

it has 51 protons, 4 more than silver and two less than xenon

Explanation:

Ca3(PO4)2 1 mole has 310.2 g

Ca3(PO4)2 4.3 × 10^-4 moles has 4.3 × 10^-4 × 310.2

= 1,333.86 × 10^-4

= 1.334 × 10^-1

OR 0.1334 g

Answer:0.178 moles

Explanation: carbon trihydride seems to be an unusual name for the methyl group CH3–

ionic wt 15

moles = 2.67/15 = 0.178

Answer:

Explanation:

1 mol of any chemical is 6.02 * 10^23

x mol of any chemical is 6.31 * 10^24

You should look at this before you put it in  your calculator. You should think roughly 10.

Now you can set it up as a proportion

1/x = 6.02 * 10^23/6.31 * 10^24        Cross multiply

6.02 * 10^23 x = 1 * 6.31 * 10^24      Divide by 6.02 * 10^23

x = 6.31 * 10^24 / 6.02 * 10^23

x = 10.48 mols.

Answer:

The kinetic theory of matter can be used to explain how solids, liquids and gases are interchangeable as a result of increase or decrease in heat energy. If it is cooled the motion of the particles decreases as they lose energy.

Explanation:

Answer:

1.023 J / g °C

Explanation:

m = 37.9 grams

ΔT = 25.0*C

H = 969 J

c = ?

The equation relating these equation is;

H = mcΔT

making c subject of formulae;

c = H / mΔT

c = 969 J / (37.9 g * 25.0*C)

Upon solving;

c = 1.023 J / g °C

Answer:

-27 C converted to Kelvin is 246.15 Kelvin.

Explanation:

Hope this helps

Answer:

246.15K

Explanation:

Formula: (-27°)C + 273.15 = 246.15K

Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O

Given

MnO4- + Br- = MnO2 + BrO3-

Required

The half-reaction

Solution

In acidic conditions :

1. Add the coefficient

2. Equalization O atoms (add H₂O) on the O-deficient side.  

3. Equalization H atoms (add H⁺ ) on the H -deficient side.  .  

4. Equalization of charge (add electrons (e) )

5. Equalizing the number of electrons and then adding the two half -reactions together

Oxidation : Br- → BrO3-  

Reduction : MnO4- → MnO2

Br- + 3H2O → BrO3-  

MnO4- → MnO2 + 2H2O

Br- + 3H2O → BrO3- + 6H+  

MnO4- + 4H+ → MnO2 + 2H2O

Br- + 3H2O → BrO3- + 6H+ + 6e-         x1

MnO4- + 4H+ + 3e- → MnO2 + 2H2O x2

Br- + 3H2O → BrO3- + 6H+ + 6e-  

2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O

Br- + 2MnO4- + 3H2O + 8H+ + 6e- → BrO3- + 2MnO2 + 6H+ + 4H2O + 6e-

Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O

The given question is incomplete. The complete question is:

How much heat is produced when 24.8 g of [tex]CH_4[/tex] is burned in excess oxygen gas

Given: [tex]CH _4 +2O_2\rightarrow CO_2+2H_2O[/tex]  ΔH= −802 kJ.

Answer: 1243.1 kJ

Explanation:

Heat of combustion is the amount of heat released on complete combustion of 1 mole of substance.

Given :

Amount of heat released on combustion of  1 mole of methane = 802 kJ kJ/mol

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

1 mole of [tex]CH_4[/tex] weighs = 16 g  

Thus we can say:  

16 g of [tex]CH_4[/tex] on combustion releases heat = 802 kJ

Thus 24.8 g of [tex]CH_4[/tex] on combustion releases =[tex]\frac{802}{16}\times 24.8=1243.1kJ[/tex]

Thus heat released when 24.8 g of methane is burned in excess oxygen gas is 1243.1 kJ

Answer:

A reflected ray always comes off the surface of a material at an angle equal to the angle at which the incoming ray hit the surface. In physics, you'll hear this called the law of reflection. You've probably heard this law stated as "the angle of incidence equals the angle of reflection

Explanation:

Answer: both

Explanation:

Answer:

[tex]1.936\times 10^{-18}\ \text{J}[/tex]

Explanation:

[tex]R_h[/tex] = Rydberg constant = [tex]2.178\times 10^{-18}\ \text{J}[/tex]

[tex]n_i[/tex] = Initial shell = 3

[tex]n_f[/tex] = Final shell = 1

We have the relation

[tex]\Delta E=R_h(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})\\\Rightarrow \Delta E=2.178\times 10^{-18}(\dfrac{1}{1^2}-\dfrac{1}{3^2})\\\Rightarrow \Delta E=1.936\times 10^{-18}\ \text{J}[/tex]

The energy of the photon emitted here is [tex]1.936\times 10^{-18}\ \text{J}[/tex].

Answer:

Your ability to move and lift objects refers to muscular strength. It's measured by how much force you can exert for a short period of time and how much weight you can lift.

Explanation:

Answer: 6.75 g

Explanation:

Answer:

1. c 2. b that's what i think i learned this in middle school i dont completely remember

Answer:

Explanation:

.

Answer:

7.08 L

Explanation:

From the question given above, the following data were obtained:

Initial mole (n₁) = 1.2 moles

Initial volume (V₁) = 5 L

Final mole (n₂) = 1.2 + 0.5 = 1.7 moles

Final volume (V₂) =?

From ideal gas equation,

PV = nRT

Divide both by n

PV / n = RT

Divide both side by P

V / n = RT/P

RT/P => constant

Therefore,

V₁ / n₁ = V₂ / n₂

With the above formula i.e

V₁/n₁ = V₂/n₂

We can obtain the final volume of the gas as follow:

Initial mole (n₁) = 1.2 moles

Initial volume (V₁) = 5 L

Final mole (n₂) = 1.7 moles

Final volume (V₂) =?

V₁/n₁ = V₂/n₂

5 / 1.2 = V₂ / 1.7

Cross multiply

5 × 1.7 = 1.2 × V₂

8.5 = 1.2 × V₂

Divide both side by 1.2

V₂ = 8.5 / 1.2

V₂ = 7.08 L

Therefore, the final volume of the gas is 7.08 L

Given :

Number of He atoms, [tex]n = 3.4\times 10^{24}[/tex] atoms.

To Find :

How many grams are their in given number of He atoms.

Solution :

We know, molecular mass of He is 4 g. It means that their are [tex]6.022 \times 10^{23}[/tex] atoms in 4 g of He.

Let, number of gram He in [tex]3.4\times 10^{24}[/tex] atoms is x , so :

[tex]x = \dfrac{3.4\times 10^{24}}{6.022\times 10^{23}}\times 4\\\\x = $$22.58\ g[/tex]

Therefore, grams of He atoms is 22.58 g .