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Tomtit [17]
3 years ago
14

Susan made observations about outside events she noticed throughout the day. The day began with rain. Susan saw puddles on the r

oad. The rain stopped and the sun began to shine. The puddles disappeared. Later, clouds began to form in the sky. How do the parts of the water cycle relate to the events that Susan saw?
Chemistry
1 answer:
S_A_V [24]3 years ago
4 0

Answer:

Rainfall - precipitation

disappeared puddles - evaporation

cloud formation - condensation

Explanation:

Rainfall that is observed by the Susan is the precipitation of the water cycle in which the water vapor that was condensed become heavy and form droplets of water and fall from sky to the earth surface.

Puddles that formed due to rainfall are the collection of water in the water cycle which is evaporated (process: evaporation) into the atmosphere n the form of water vapor which condensed to form clouds.

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Which two organelles are commonly found in plant cells, help to protect the cell as well as make food, and are NOT found in anim
vampirchik [111]

Answer:

this is 6th grade work

Explanation:

4 0
3 years ago
It was calculated that 4.3mL of 0.417 M HCl is required to titrate 11.9 mL of 0.151 M Mg(OH)2. Show evidence 2 HCl(aq) + Mg(OH)2
Lapatulllka [165]

Answer:

See explanation.

Explanation:

Hello,

In this case, for the described chemical reaction:

2 HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2 H2O(l)

We can notice there is a 2:1 molar ratio between the moles of hydrochloric acid and magnesium hydroxide, therefore, at the equivalence point:

n_{HCl}=2*n_{Mg(OH)_2}

And in terms of volumes and concentrations we verify:

V_{HCl}M_{HCl}=2*V_{Mg(OH)_2}M_{Mg(OH)_2}

So we use the given data to proof it:

4.3mL*0.417M=2*11.9mL*0.151M\\1.793=3.594

Therefore, we can conclude the data is wrong by means of the 2:1 mole ratio that for sure was not taken into account. This is also supported by the fact that normalities are actually the same, but the nomality of magnesium hydroxide is the half of the hydrochloric acid normality since the acid is monoprotic and the base has two hydroxyl ions.

Best regards.

4 0
3 years ago
A sample of sodium reacts completely with 0.497 kg of chlorine, forming 819 g of sodium chloride. what mass of sodium reacted?
Slav-nsk [51]

Sodium reacts to chlorine and gives NaCl. The balanced reaction is given below:

2Na + Cl₂→ 2NaCl. Two moles Na reacts with one mole Cl₂ and produces two moles of NaCl. Atomic mass of Na= 23, Molar mass of Cl₂= 71, molar mass of NaCl=58.5.

So, 46 g Na reacts with 71 g of Cl₂ and produces (2 X 58.5)g = 117 g of NaCl. As per question Na reacts completely which means Na is the limiting reagent. So, number of moles of Na reacts = number moles of NaCl produced.

NaCl produced= 819 g= (819/58.5) moles= 15.69 moles. Therefore, 15.69 moles = 15.69 X 23 g=360.87 g of Na reacted.

8 0
3 years ago
The Sun heats land and water, how is this related to wind and ocean currents?
Yuki888 [10]

Answer:

"The sun warms up parts of the oceans. Warm waters rise just like warm air rises. So, as the warmer ocean waters begin to rise in a particular area, the cooler ocean waters from a different area will move in to replace the warmer ocean waters, and this creates our ocean currents."

Explanation:

Hope this is helpful :)

8 0
3 years ago
What is the boiling point of water when 175.0 g of Na2SO4, a strong electrolyte is dissolved in 1.000 Kg of water?
liubo4ka [24]

Answer: 101.9^0C

Explanation:

Elevation in boiling point is given by:

\Delta T_b=i\times K_b\times m

\Delta T_b=T_b-T_b^0=(T_b-100)^0C = Elevation in boiling point

i= vant hoff factor = 3 (number of ions an electrolyte produce on complete dissociation)

Na_2SO_4\rightarrow 2Na^++SO_4^{2-}

K_f = freezing point constant = 0.512^0C/m

m= molality

\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (water)= 1.000 kg

Molar mass of solute Na_2SO_4 = 142 g/mol

Mass of solute Na_2SO_4  = 175.0 g

(T_b-100)^0C=3\times 0.512\times \frac{175.0g}{142g/mol\times 1.000kg}

T_b=101.9^0C

Thus the boiling point of water when 175.0 g of Na_2SO_4, a strong electrolyte is dissolved in 1.000 Kg of water is 101.9^0C

8 0
3 years ago
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