3. Ni(NO2)3
What’s the name for the formula

Answer:

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

Explanation:

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:

antimony

Explanation:

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

Answer:

See explanation

Explanation:

The greater the relative atomic  mass of an atom the more polarizable the atom is.

Also, the magnitude of dispersion forces depends on the molecular mass of the compound.

We must remember that both I2 and F2 are nonpolar molecules that are purely held together by dispersion forces.

Hence, I2, having the greater molecular mass and surface area experiences a greater magnitude of dispersion forces and consequently has a higher boiling point than F2.

Answer: 6.75 g

Explanation:

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:

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

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

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

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:

-27 C converted to Kelvin is 246.15 Kelvin.

Explanation:

Hope this helps

Answer:

246.15K

Explanation:

Formula: (-27°)C + 273.15 = 246.15K

Answer:

C. Ionic-Li and S are two nonmetals

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].

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 .

Answer:

SrO = Sr + O - Chemical Equation Balancer.

Explanation:

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:

*it can cause the atom to become and ion

Answer:

I hope I can help

Explanation:

I hope I can help you

Answer: both

Explanation:

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:

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:

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

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.