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3 years ago

What is the product in the reaction below ??

Chemistry

1 answer:

kupik [55]3 years ago

4 0

A. ZnCl2 + 2 Ag → 2 AgCl2 + Zn

B. NaF + K → Na + KF

C. AgCl + Cu → CuCl + Ag

D. Fe2O3 + C → Fe + CO2

E. Fe2O3 + Al → Fe + Al2O3

Hope it helped.. pls mark brainliest

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A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by th

Ann [662]

Answer:

The ΔH of the reaction is + 12.45 KJ/mol

Explanation:

Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)

heat capacity of water = 4.18 Jk-1 Mol-1

Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)

Molar mass of NaHCO3 = 84 g/mol

Mole of NaHCO3 = 14.5 / 84 = 0.173 mol

Step 1 : Calculate the heat energy (Q) lost by the water.

Q = M x C x ΔT

Q = -100 x 4.18 x (-5.14)

Q = 2148.5 joules

Q = 2.1485 K J

Step 2: Calculating the ΔH of the reaction?

ΔH = Q / number of moles of NaHCO3

ΔH = 2.1485 / 0.173

ΔH = 12.42 KJ/mol

3 0

3 years ago

7.In which state of matter molecules have more energy to vibrate.

viva [34]

Answer:

Solid, liquid or gas. So there is movement no matter the state. The key variable is density. The higher the density the less movement. In solids the motion can be so small it's very hard to measure. Gas on the other hand is easy the motion being large. Bear in mind temperature plays a big role. Higher temps bring faster motion. Finally the pressure of the gas brings about less motion the higher it is as the molecules are closer together & can't move as much.

Explanation:

8 0

3 years ago

The normal boiling point for acetone is 56.5°c. at an elevation of 5300 ft the atmospheric pressure is 630. torr. what would be

Flauer [41]

20.5 because the boiling point should decrease not increase

7 0

3 years ago

If the energy difference between two electronic states is 214.68 kJ / mol , calculate the frequency of light emitted when an ele

atroni [7]

${\qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

Energy difference btween the two electronic states can be expressed as :

${ \qquad \sf \dashrightarrow \: \Delta E = h\nu}$

[ h = planks constant, ${\: \nu }$ = frequency ]

$\qquad \sf \dashrightarrow \:214.68 = 39.79 \times 10 {}^{ - 14} \times \nu$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 \times 10 {}^{ - 4} }$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 } \times 10 {}^{14}$

$\qquad \sf \dashrightarrow \: \nu \approx 5.395 \times10 {}^{14} \:\:hertz$

5 0

2 years ago

In another experiment, a 0.150 M BF4^-(aq) solution is prepared by dissolving NaBF4(s) in distilled water. The BF4^-(aq) ions in

Ilia_Sergeevich [38]

Answer:

A) Forward rate = 1.1934 × 10^(-4) M/min

B) I disagree with the claim

Explanation:

A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.

The reversible reaction given to us is;

BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

From this, we can see that the stoichiometric ratio is 1:1:1:1

Thus, concentration of [BF4-] is now;

[BF4-] = 0.150 - 0.0174

[BF4-] = 0.1326 M

From the rate law, we are told the forward rate is kf [BF4-].

We are given Kf = 9.00 × 10^(-4) /min

Thus;

Forward rate = 9.00 × 10^(-4) /min × (0.1326M)

Forward rate = 1.1934 × 10^(-4) M/min

(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.

Thus, I disagree with the claim.

3 0

3 years ago