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Sveta_85 [38]
3 years ago
10

Please help time is of the essence:

Chemistry
1 answer:
Juliette [100K]3 years ago
4 0
The Indicator changes color in a titration is Indicator Point 
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Which statement correctly pairs the climate factor with its effect on temperature? *
allochka39001 [22]

Answer:

A region on top of a mountain is cooler than at the base.

Explanation:

Pressure and temperature have direct relationship with each other. With the decrease in pressure, the temperature decreases and vice versa. When the air rises in the atmosphere, the pressure starts to fall. The low pressure at the peak of the mountains tends to cause the fall in temperature. It is because of this reason that it is cooler at the top of the mountain while the temperature is less cool in the foothills.

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2 years ago
How many grams are contained in a 0.183 mol sample of ammonium phosphate?
soldi70 [24.7K]

Molar mass of ( NH₄)₃PO₄ = 14.01×3 + 1.01×12 + 30.97 + 16.00×4 = 149.12 g/mol. Mass of 0.183 mol ...

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2 years ago
In terms of their electron configurations, why is cesium more likely to lose its valence electron than potassium?
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Explanation:

use the term electron sheilding, the more electrons between the valence el3ctron and nucleus the easier to lose the valence electron (more sheilding = easier to lose)

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3 years ago
PLEASE HELP TEST<br> does anyone know the answer???? density question
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The answer is honey.

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8 0
2 years ago
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The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C
Kipish [7]

Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

7 0
3 years ago
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