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LenaWriter [7]
3 years ago
14

Study the image. Warm and cold air fronts with points labeled 1 through 4. 1: cold air entering from the left and moving down. 2

: clouds with rain. 3: warm air entering from the right and moving up. 4: warm air entering from the top left and moving down toward the clouds. Warm and cold air fronts with points labeled 1 through 4. 1: cold air entering from the left and moving down. 2: clouds with rain. 3: warm air entering from the right and moving up. 4: warm air entering from the top left and moving down toward the clouds. At which point is cool air circulating beneath warm air? 1 2 3 4
Chemistry
2 answers:
nata0808 [166]3 years ago
7 0

Answer:

1

Explanation:

NISA [10]3 years ago
4 0

Answer:

1

Explanation:

Edge 2021

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If an organism is compose of only one cell, it ____.
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C or A I may be incorrect but those are what I think it is

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Explanation:

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A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.74 g of water at 52.2 oC in an insulated container. clead = 0.128
QveST [7]

Answer:

Explanation:

Hello, in this case, the lead is catching heat and the water losing it, that's why the heat relation ship is (D is for Δ):

DH_{lead}=-DH_{water}

Now, by stating the heat capacity definition:

m_{Pb}C_{Pb}*(T_{eq}-T_{lead}=-m_{H_2O}C_{H_2O}*(T_{eq}-T_{H_2O})\\

Solving for the equilibrium temperature:

T_{eq}=\frac{m_{Pb}C_{Pb}T_{Pb}+m_{H_2O}C_{H_2O}T_{H_2O}}{m_{Pb}C_{Pb}+m_{H_2O}C_{H_2O}} \\\\T_{eq}=\frac{2.04g*0.128J/(g^oC)*10.8^oC+7.74g*4.18J/(g^oC)*52.2^oC}{2.04g*0.128J/(g^oC)+7.74g*4.18J/(g^oC)} \\\\T_{eq}=51.87^oC

Which is very close to the water's temperature since the lead's both mass and head capacity are lower than those for water.

Best regards.

5 0
2 years ago
A world without friction would it be?<br> Bill Nye Video
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8 0
2 years ago
4. Calculation of theoretical yield and percent yield You balanced this reaction earlier: Mg(s) + HCl (aq) H2 (g) + MgCl2 (aq) Y
erik [133]

Answer:

Mg(s) +<em> 2</em> HCl (aq) →  H₂(g) + MgCl₂

0.415g of H₂(g) <em>-Assuming mass of Mg(s) = 10.0g-</em>

Explanation:

Balancing the reaction:

Mg(s) + HCl (aq) →  H₂(g) + MgCl₂

There are in products two atoms of H and Cl, the balancing equation is:

Mg(s) +<em> 2</em> HCl (aq) →  H₂(g) + MgCl₂

<em>Assuming you add 10g of Mg(s) -Limiting reactant-</em>

<em />

10g of Mg are (Atomic mass: 24.305g/mol):

10g × (1 mol / 24.305g) = <em>0.411 moles of Mg</em>

<em>-Theoretical yield is the amount of product you would have after a chemical reaction occurs completely-</em>

Assuming theoretical yield, as 1 mole of Mg(s) produce 1 mole of H₂(g), theoretical yield of H₂(g) is 0.411moles H₂(g). In grams:

0.411mol H₂(g) × (1.01g / mol) = <em>0.415g of H₂(g)</em>

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2 years ago
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