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3 years ago

tres pensamientos negativos sobre el C0Vid-19 que se vive en la actualidad y luego transformemos a pensamientos positivos.

Chemistry

2 answers:

katrin2010 [14]3 years ago

7 0

Answer:

Bueno,

Explanation:

Tres pesnamientos negativos sobre el C0Vid-19 son,

- Encerrados en la casa

- Tener que ponerse mascara

- Todos los lugares divertidos cerrados

Convertidos en positivos:

- Pasamos mas tiempo de calidad en familia.

- Nos protejemos unos a otros.

- Descansan un poco todas las personas que se han pasado toda la vida trabajando.

iragen [17]3 years ago

6 0

Answer:

estamos encerrados pero ahora el ecosistema esta tomando forma como antes

es muy contagiable pero si nos contagiamos, nuestro cuerpo forma la imunidad en ese virus y hasta en algunos parecido

no puedo ver a mis amigos pero puedo pasar mas tiempo y disfrutar a mi familia

Que tengas buen dia! :)

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Answer:

The correct answer is B.

Explanation:

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The ratio of masses are given as:

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This illustrates the law of definite proportions which is also known as law of constant compositions .

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Whereas :

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3 years ago

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Arte-miy333 [17]

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2 years ago

Read 2 more answers

A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$