Based on current information, cosmologists expect the universe to _____.

1 answer:

8 0

They expect the Universe to continue expanding. Right now, the evidence is that the expansion is speeding up. They also expect entropy to continue increasing, so in the distant future, there will not be any free energy left to support life, for example.

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What is the net ionic equation for the reaction if any that occurs when aqueous solutions of Na2CO3 and HCL are mixed?

strojnjashka [21]

Answer:

CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)

Explanation:

The balanced reaction between Na2CO3 and HCl is given as;

Na₂CO₃ (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO₂ (g) + H₂O (l)

The next step is o express the species as ions.

The complete ionic equation for the above reaction would be;

2Na⁺(aq) + CO₃²⁻(aq) + 2H⁺(aq) + 2Cl⁻(aq) → Na⁺(aq) + Cl⁻(aq) + CO₂ (g) + H₂O (l)

The next step is to cancel out the spectator ion ions; that is the ions that appear in both the reactant and product side unchanged.

The spectator ions are; Na⁺ and Cl⁻

The net ionic equation is given as;

CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)

7 0

3 years ago

If the cylindrical pistons are 25.000 cm in diameter at 20.0 ∘c, what should be the minimum diameter of the cylinders at that te

attashe74 [19]

Refer to the diagram shown below.

The piston supports the same load W at both temperatures.
The ideal gas law is
$pV=nRT$
where
p = pressure
V = volume
n = moles
T = temperature
R = gas constant

State 1:
T₁ = 20 C = 20+273 = 293 K
d₁ = 25 cm piston diameter

State 2:
T₂ = 150 C = 423 K
d₂ = piston diameter

Because V, n, and R remain the same between the two temperatures, therefore
$\frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}}$

If the supported load is W kg, then
$p_{1} = \frac{W \, N}{ \frac{\pi}{4} d_{1}^{2}} = \frac{4W \, N}{\pi (0.25 \, m)^{2}} = 20.3718W \, Pa$
Similarly,
$p_{2} = \frac{4W}{\pi d_{2}^{2}} \, Pa$

$\frac{p_{1}}{p_{2}} = \frac{20.3718 \pi d_{2}^{2}}{4} = 16 d_{2}^{2}$

Because p₁/p₂ = T₁/T₂, therefore
$16d_{2}^{2} = \frac{293}{423} \\\\ d_{2}^{2} = \frac{0.6927}{16} \\\\ d_{2} = 0.2081 \, m$

The minimum piston diameter at 150 C is 20.8 cm.

Answer: 20.8 cm diameter