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2 years ago

What’s chemical formula ammonium iodide?

Chemistry

2 answers:

Law Incorporation [45]2 years ago

8 0

Answer:

NH4I

..................

I hope it helps

Alexeev081 [22]2 years ago

4 0

Answer:Ammonium Iodide is a chemical compound with a formula Explanation: nh4l

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If the starting volume of a hot air balloon is 55,500 m3and the initial temperature is 21 °C, what is the temperature inside the

Dmitry_Shevchenko [17]

Answer:

T₂ = 392 K

Explanation:

Given that,

Initial volume of the hot air balloon, V₁ = 55500 m³

Initial temperature, T₁ = 21°C = 294 K

Final volume, V₂ = 74000 m³

We need to find the final temperature inside the balloon. The relation between the temperature and volume is given by charles law i.e.

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Where

T₂ is the final temperature

So,

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{294\times 74000 }{55500 }\\\\T_2=392\ K$

So, the new temperature is 392 K.

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3 years ago

How many moles of mercury, Hg, are there in 1.30 x 10^7 atoms of murcury? PLEASE HELP, IT'S URGENT! Thank you! (:

vodka [1.7K]

1 mole Hg ---------------- 6.02x10²³ atoms
?? ------------------------- 1.30 x10⁷ atoms

1.30x10⁷ x 1 / 6.02x10²³ =

= 1.30x10⁷ / 6.02x10²³ => 2.159x10⁻¹⁷ moles

hope this helps!

7 0

3 years ago

You have 1 mole of a gas at STP. If you apply the ideal gas law what is the approximate volume of the gas?

goldfiish [28.3K]

Answer:

A) 22.4L

Explanation:

we know, ideal gas law states

PV=nRT

V=nRT/P

At STP,

T= 273.15K P=1atm R=0.082L.atm/mol/K n=1 mole

V=(1*0.082*273.15)/ 1

V=22.4L

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3 years ago

What is the mass of 1.84 mol NaCl?

Goshia [24]

Sent a pic that shows the work and answer is in box.

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3 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0

4 years ago