The two south pole ends of two magnets are touching. Which of the following can be concluded?

How many moles are there in 545g of <img src="https://tex.z-dn.net/?f=" id="TexFormula1" title="" alt="" align="absmiddle" class

Pachacha [2.7K]

answer: 3.40625 moles

Explanation: see the attached pics

8 0

3 years ago

How much heat is evolved when 22.5 grams of a food sample changes the temperature of 0.42 kg of water by 4.4 degree C. Assume sp

irakobra [83]

The answer is C. Assume specific heat to be 4.18 J/g/C

3 0

3 years ago

How TO DO THIS QUESTION PLEASE PLEASEEEEEEEEEeeeeeeeeeeeeeeeeeeee

krok68 [10]

O2 is the limited reactant

3 0

3 years ago

Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride

GalinKa [24]

Answer:

HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)

Explanation:

Complete question

Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride. What is the net ionic equation

Equilibrium equation between the undissociated acid and the dissociated ions

HF(aq)⇌H+(aq)+F−(aq)

Sodium hydroxide will dissociate aqueous solution to produce sodium cations, Na+, and hydroxide anions, OH−

NaOH(aq)→Na+(aq)+OH−(aq)

Hydroxide anions and the hydrogen cations will neutralize each other to produce water.

H+(aq)+OH−(aq)→H2O(l)

On combining both the equation, we get –

HF(aq)+Na+(aq)+OH−(aq)→Na+(aq)+F−(aq)+H2O(l)

The Final equation is

HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)

5 0

3 years ago

Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c

uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

$I_{2} (g) + Cl_{2} (g) -----> 2 ICl (g)$

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

$K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}$

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = $\frac{(2x)^{2}}{(0.437-x) (0.269-x)}$

81.9 = $\frac{(2x)^{2}}{x^{2} -0.706x + 0.118}$

$(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]$

$4x^{2} = 81.9x^{2} -57.8x +9.66$

$77.9x^{2} -57.8x+9.66 = 0$

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

$[ICl]_{eq} = 2 ( 0.254) = 0.508 M$

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0

3 years ago