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balandron [24]
2 years ago
10

A lahar is a type of mudflow that occurs?

Chemistry
2 answers:
mixas84 [53]2 years ago
5 0

A lahar is a type of mud flow that occurs?  The answer is, A. after a volcanic eruption.

Natalka [10]2 years ago
4 0
<span>A lahar is a type of mudflow that occurs after a volcanic eruption. It is a sort of mudflow or trash stream made out of a slurry of pyroclastic material, rough flotsam and jetsam, and water. The material streams down from a fountain of liquid magma, normally along a waterway valley. Lahar is the most deadly by-products of the eruption because of the speed they can travel. </span>
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¿Qué compone a un ecosistema?
schepotkina [342]

An ecosystem is a geographic area where plants, animals, and other organisms, as well as weather and landscape, work together to form a bubble of life. Ecosystems contain biotic or living, parts, as well as abiotic factors, or nonliving parts. Biotic factors include plants, animals, and other organism .

8 0
3 years ago
Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2
baherus [9]

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) Cu_{(s)} →Cu^{2+}_{(aq)} +2e E°=0.337 v

(reduction) MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}→MnO_{2 (aq)}+2H_{2}O E°=1.679 v

(overall) 2MnO_{4 (aq)}+3Cu_{(s)}+8H^{+}_{(aq)}→3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346

E=1.346

5 0
2 years ago
Which statement is true about matter?
sergeinik [125]

Answer:

C

Explanation:

This is because matter is anything that has mass and occupies space.Therefore the space occupied by matter is volume

6 0
2 years ago
Two systems with heat capacities 19.9 J mol-1 K-1 and 28.2 ] mol 1 K-1 respectively interact thermally and come to an equilibriu
MAVERICK [17]

Answer : The initial temperature of system 2 is, 19.415^oC

Explanation :

In this problem we assumed that the total energy of the combined systems remains constant.

-q_1=q_2

m\times c_1\times (T_f-T_1)=-m\times c_2\times (T_f-T_2)

The mass remains same.

where,

C_1 = heat capacity of system 1 = 19.9 J/mole.K

C_2 = heat capacity of system 2 = 28.2 J/mole.K

T_f = final temperature of system = 30^oC=273+30=303K

T_1 = initial temperature of system 1 = 45^oC=273+45=318K

T_2 = initial temperature of system 2 = ?

Now put all the given values in the above formula, we get

-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K

T_2=292.415K

T_2=292.415-273=19.415^oC

Therefore, the initial temperature of system 2 is, 19.415^oC

8 0
3 years ago
Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.
Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0
2 years ago
Read 2 more answers
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