Sulfurous acid reacts with barium hydroxide.

Is The reaction time is different for different stimuli true or false?

MArishka [77]

Answer: true

Explanation: In this example, the processes (perceive, process, and respond), are done in a matter of milliseconds, but reaction time can vary depending on a variety of factors: Complexity of the stimulus-The more complex the stimulus, the more information that has to be processed, the longer this process will take.

5 0

3 years ago

How does the vegetation surface type affect the amount of runoff

maxonik [38]

Answer:

The type of vegetation a surface does affect the water coming from above to sink in or runoff. 

Explanation:

This is how the vegetation affects the runoff:-

The leaves and stems present in the vegetation do not let the water fall directly on the soil and makes the process rather slow which makes the water to get to the ground slowly and sink in properly inside the soil rather than running off.

If the vegetation present is dense with there was being hairy then also the water would not run out and will get absorbed by the roots letting the soil intact

5 0

2 years ago

The temperature of a sample of matter is a measure of the ?

Masteriza [31]

Answer: the average kinetic energy of the particles.


Jusitification:


Temperature and heat energy are closely related.


While heat is the kinetic energy of the particles of a substance which is transferred from a hotter substance to a cooler one, the temperature is a measure of the average kinetic energy of the particles in a substance.


The relatively high kinetic energy the particles of a warm substance is transferred to the cooler one by the motion (vibration or translation) of the atoms of molecules. The energy transferred is heat energy.

3 0

3 years ago

PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Masteriza [31]

Answer:

$\large \boxed{\text{-827.4 kJ}}$

Explanation:

We have three equations:

1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ

3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.

7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

6. 2H₂S(g) + O₂(g) ⟶ 2S(s) + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

$\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}$

8 0

3 years ago

Why do real gases deviate from gas laws

frosja888 [35]

Answer:

Attractions between molecules cause a reduction in volume

Explanation:

5 0

2 years ago