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slavikrds [6]
3 years ago
11

Which method used to attack ground cover fires is generally used against fires that are very hot, very big, and very fast?

Chemistry
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

Indirect attack (ground cover)

Explanation:

Indirect attack (ground cover) is a method of controlling a groundcover fire whereby the control line is erected or situated at a remote from the main fire edge, and the fuel between the two points was burned.

The indirect attack is set up at some distance

from the raging fire. The indirect attack method is normally used to put out

fires that are too fast, too big or too hot for a direct attack method to work.

adjustments can be made when required

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A chemist measures the energy change ΔH during the following reaction: C3H8 (g) +5O2 (g) →3CO2 (g) +4H2O (l) =ΔH−2220.kJ Use the
statuscvo [17]

Answer:

The reaction is exothermic.

Yes, released.

The heat released is 4,08x10³ kJ.

Explanation:

For the reaction:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

The ΔH is -2220 kJ, As ΔH is <0, <em>The reaction is exothermic.</em>

As the reaction is exothermic, the heat of the reaction will be <em>released.</em>

The heat released in 81,0g is:

81,0g C₃H₈×\frac{1mol}{44,1g}×\frac{2220kJ}{1mol}= <em>4,08x10³ kJ</em>

<em>-Using molar mass of C₃H₈ to convert mass to moles and knowing that there are released 2220 kJ per mole of C₃H₈-</em>

I hope it helps!

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3 years ago
How does the law of conservation of mass relate to the number of atoms of each element that are present before a reaction vs. th
TiliK225 [7]
The law of conservation of mass or principle of mass conservation states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as system's mass cannot change, so quantity cannot be added nor removed. Hence, the quantity of mass is conserved over time.

The law implies that mass can neither be created nor destroyed, although it may be rearranged in space, or the entities associated with it may be changed in form. For example, in chemical reactions, the mass of the chemical components before the reaction is equal to the mass of the components after the reaction. Thus, during any chemical reaction and low-energy thermodynamic processes in an isolated system, the total mass of the reactants, or starting materials, must be equal to the mass of the products.

According to the Law of Conservation, all atoms of the reactant(s) must equal the atoms of the product(s).
As a result, we need to balance chemical equations. We do this by adding in coefficients to the reactants and/or products. The compound(s) itself/themselves DOES NOT CHANGE.
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3 years ago
What property of water allows it to stick to a dry surface
son4ous [18]
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4 0
4 years ago
Mercury thermometers work because mercury ____ when it is warmed
kap26 [50]
Mercury expands when it is heated. This process is called thermal expansion.
5 0
3 years ago
Consider the following system at equilibrium:
alexgriva [62]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

6 0
3 years ago
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