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balandron [24]
3 years ago
10

A lahar is a type of mudflow that occurs?

Chemistry
2 answers:
mixas84 [53]3 years ago
5 0

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

Natalka [10]3 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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Hi! I need help with this question, if it is possible, please answer immediately.
Dennis_Churaev [7]

Answer:

A.

Explanation:

The paper clip allows elictricity to pass, unlike the eraser or paper

6 0
3 years ago
Guys help me out please ​
kifflom [539]

Answer:

See explanation

Explanation:

A net ionic equation shows only the chemical species that are involved in a reaction, while a complete ionic equation also includes the spectator ions(Khan Academy).

From the species shown in the image we can write the molecular equation, ionic equation and net ionic equation and also identify the spectator ions.

Chemical equation: KX + BY --------> BX + AY

Ionic Equation; X^- + A^+ + B^+ + Y^- -----> BX + A^+ + Y^-

Spectator ions;  A^+ and Y^-

Net Ionic Equation; X^- + B^+ ------> BX

7 0
2 years ago
When 10.0 grams of ch4 reacts completely with 40.0 grams of o2 such that there are no reactants left over, 27.5 grams of carbon
Zina [86]

The balanced equation for the reaction is :-

CH₄(g) + 2O₂(g)  ---------> CO₂(g) + 2H₂O(l)

Molar mass of CH₄ = 16 g/mole

Molar mass of O₂ = 32 g/mole

Molar mass of H₂O = 18 g/mole

Molar mass of CO₂ = 44 g/mole

Now we calculate the number of moles of reactants,

moles of CH₄ = mass of CH₄/molar mass of CH₄ = 10/16 = 0.625

Moles of O₂ = mass of O₂/molar mass of O₂ = 40/32 = 1.25

Now, as per the balanced reaction, for complete reaction to occur, one mole of CH₄ require 2 moles of O₂

Thus, 0.625 moles of CH₄ requires 1.25 moles of O₂

both CH₄ and O₂ are present in the exact required quantity.

Hence, moles of CO₂ formed = moles of CH₄ reacted = 0.625

Mass of CO₂ formed = moles of CO₂ x Molar mass of CO₂ = 0.625 x 44 = 27.5g

Moles of H₂O formed = moles of O₂ reacted = 1.25g

<span>Thus, mass of H</span>₂O formed = moles of H₂O x molar mass of H₂O = 1.25 x 18 = 22.5g

8 0
2 years ago
Propane (C3H8) can be burned to produce heat for homes. The products of the reaction are CO2 and H2O. For complete combustion to
Natalija [7]

Answer:

1- 3 Moles of CO2  

2- 132 g of CO2  

3- 105,6 g of CO2

4- Limiting Reagent O2

<u>Products form based on limiting reagent (384g O2) :</u>

CO2: 316,8 g

H2O: 172,8 g

<u>Products form based on C3H8 (132,33 g):</u>

CO2: 396,99 g

H2O: 216,54 g  

Explanation:

<u>Atomic Masses:</u>

C: 12

H: 1

O: 16

<u>Molecular weights:</u>

C3H8: 44 g

O2: 32 g

H2O: 18 g

CO2: 44 g

C3H8 + 5 O2⇒ 3 CO2 + 4 H2O

C3H8 (44g)+ O2 (160 g) ⇒ CO2 (132 g) + H2O (72 g)

In 5 moles of O2 are produced 3 moles of CO2, equivalent to 132 g

For 160g of O2 are produced 132 g of CO2, so 128 g of O2

160 g  O2 ⇒ 132 g CO2

128 g  O2⇒ × = 105,6 g CO2

(128×132÷160= 105,6)

The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.

If I have 132,33 g of C3H8 and 384 g O2 we can calculate:

For          44 g of CH3H8  ⇒160 g of O2

With   132,33 g of CH3H8 ⇒ ×= 481,2 g of O2

(132,33×160÷44=481,2)

As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.

<u>Calculations of the products formed in base of quantity of O2 (limiting reagent):</u>

160 g  O2 ⇒ 132 g CO2

384 g  O2⇒ × = 316,8 g CO2

(384×132÷160= 316,8)

160 g  O2 ⇒ 72 g H2O

384 g  O2⇒ × = 172,8 g H2O

(384×72÷160= 172,8)

<u>Calculations of the products formed in base of quantity of C3H8 (excess reagent):</u>

     44 g  C3H8 ⇒ 132 g CO2

132,33 g  C3H8 ⇒ × = 396,99 g CO2

(132,33×132÷44=396,99)

     44 g C3H8 ⇒ 72 g H2O

132,33 g  C3H8⇒ × = 216,54 g H2O

(132,33×72÷44= 216,54)

<u />

3 0
3 years ago
Which of the following compounds is SOLUBLE? <br> A. SrSO4 <br> B. CaCO3 <br> C. BaS <br> D. CaCl2
andriy [413]

Answer:

A).

Strontium Sulphate is soluble

3 0
3 years ago
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