Answer:
1.001
Explanation:
The Significant Figures are 1 0 0 1, This answer has 4 Significant figures, while the other three have only 2 significant figures
Answer: The equilibrium concentration of hydrogen gas is 0.0269 M
Explanation:
The chemical reaction follows the equation:

At t = 0 0.044M 0.044M 0.177M
At
(0.044-x)M (0.044-x)M (0.177+x)M
The expression for
for the given reaction follows:
![K_c=\frac{[HI]^2}{[H_2]\times [I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5Ctimes%20%5BI_2%5D%7D)
We are given:

Putting values in above equation, we get:


Hence, the equilibrium concentration of hydrogen gas is (0.044-x) M =(0.044-0.0171) M= 0.0269 M
Answer:
I believe it's A. to reduce air bubbles. Tbh, it's been a while
Answer:
1.2 × 10⁴ cal
Explanation:
Given data
- Initial temperature: 80 °C
We can calculate the heat released by the water (
) when it cools using the following expression.

where
c is the specific heat capacity of water (1 cal/g.°C)

According to the law of conservation of energy, the sum of the heat released by the water (
) and the heat absorbed by the reaction (
) is zero.

Answer:
5.5 atm
Explanation:
Step 1: Calculate the moles in 2.0 L of oxygen at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
2.0 L × 1 mol/22.4 L = 0.089 mol
Step 2: Calculate the moles in 8.0 L of nitrogen at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
8.0 L × 1 mol/22.4 L = 0.36 mol
Step 3: Calculate the total number of moles of the mixture
n = 0.089 mol + 0.36 mol = 0.45 mol
Step 4: Calculate the pressure exerted by the mixture
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 0.45 mol × (0.0821 atm.L/mol.K) × 298 K / 2.0 L = 5.5 atm