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Sphinxa [80]
4 years ago
10

Porque sirve el teorema de pitagora​

Chemistry
1 answer:
denis23 [38]4 years ago
4 0

El teorema de Pitágoras se generaliza a espacios de dimensiones superiores. Algunas de las generalizaciones están lejos de ser obvias. El teorema de Pitágoras sirve como base de la fórmula de distancia euclidiana.

Espero que esto ayude, que tengan un día bendecido y maravilloso, ¡así como uno seguro! :-)

-Cutiepatutie

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what is the boiling point of the solution resulted from the dissolving of 32.5g of NaCl in 250.0g of water?
suter [353]

Answer:

The boiling point of this solution is 102.28 °C

Explanation:

<u>Step 1:</u> Data given

Mass of Nacl = 32.5 grams

Molar mass of NaCl = 58.45 g/mol

Mass of Water = 250 grams

Boiling point of water = 100°C

<u>Step 2: </u>Calculate number of moles

Number of moles = mass of NaCl / Molar mass of NaCl

Number of moles = 32.5 grams /58.45 g/mol = 0.556 moles

<u>Step 3:</u> Calculate molality

Molality = Number of moles / mass of water

Molality = 0.556 moles / 0.250 kg of water

Molality = 2.224 molal

NaCl releases twice as many moles of ions, the total molality of this solution is also twice: 4.448 molal

Step 4: Calculate boiling point

dT =( 0.512 C / molal)*4.448 molal)

dT = 2.28

The boiling point of this solution is 100 °C + 2.28 °C = 102.28 °C

8 0
3 years ago
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr(aq)+
Mkey [24]

Full Question:

A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?

How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?

2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)

Answer:

13.1 g K2CO3 required to neutralize spill

Explanation:

2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)

Number of moles = Volume * Molar Concentration

moles HBr= 0.42L x .45 M= 0.189 moles HBr

From the stoichiometry of the reaction;

1 mole of K2CO3 reacts  with 2 moles of HBr

1 mole = 2 mole

x mole = 0.189

x = 0.189 / 2 = 0.0945 moles

Mass = Number of moles * Molar mass

Mass = 0.0945 * 138.21  = 13.1 g

3 0
3 years ago
Number of moles in 3.70 x 10^-1 g of boron
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The answer is 10.81 g of boron.
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