Q = mcΔt, q = energy [J] m = mass (of water) [g]; c = specific heat capacity of water [J g⁻¹ K⁻¹/°C⁻¹]; Δt = change in temperature [K/°C]
Δt = 121 - -24 = 145
q = 39 × 4.18 × 145
q = 23637.9 J

4 0

3 years ago

For the chemical equation SO 2 ( g ) + NO 2 ( g ) − ⇀ ↽ − SO 3 ( g ) + NO ( g ) the equilibrium constant at a certain temperatur

MakcuM [25]

Answer:

Moles of NO₂ = 0.158

Explanation:

SO 2 ( g ) + NO 2 ( g ) ⇄ SO 3 ( g ) + NO ( g )

According to the law of mass equation

$K_{c}$ = $\frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2} ]}$

⇒ 3.10 = $\frac{(1.00)(1.00)}{(2.30) [NO_{2} ]}$ At equilibrium [SO₃] = [NO]

⇒ [NO₂] = $\frac{1}{6.3}$

⇒ [NO₂] = 0.158

So. number of moles of NO₂ at equilibrium added = 0.158

8 0

3 years ago

In a synthesis reaction, one reactant contains 346 J of chemical energy, and one reactant contains 153 J of chemical energy. The

Eduardwww [97]

Answer:

64J of energy must have been released.

Explanation:

Step 1: Data given

One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.

The product contains 435 J of chemical energy.

Step 2:

Since the energy is conserved

Sum of energy of Reactants = Energy of Products

Sum of energy of Reactants = 346 J + 153 J = 499 J

The energy of the product = 435 J

435 < 499

This means energy must have been lost as heat.

Step 3: Calculate heat released

499 J - 435 J = 64 J

64J of energy must have been released.

4 0

3 years ago

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is $4.58\times 10^{-3}M$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

4 0

3 years ago

How scientists used the scientific method to learn more about how dinosaurs became extinct​

How scientists used the scientific method to learn more about how dinosaurs became extinct