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3 years ago

How calculate how many milliliters of glycerin (specific gravity=1.26) will have a mass of 0.75 lbs?

1 answer:

5 0

Specific gravity is the ratio between the density of the material to the density of water at 4 degrees celcius (density of water at this temperature is 1000 kg/m^3)

1.26 = density of glycerin / 1000
density of glycerin = 1260 kg/m^3

1 Kg is equivalent to 2.2 lbs, therefore:
0.75 lbs = 0.34 kg

density = mass/volume
1260 = 0.34/volume
volume = 0.34/1260 = 2.857 x 10^-4 m^3

now, 1 m^3 is equivalent to 1000 liters
2.857 x 10^-4 m^3 = 0.2857 liters

one liter is equivalent to 1000 ml
therefore,
0.2857 l = 285.7 ml

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In which of the following locations are bacteria often found?

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3 years ago

The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The

dsp73

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

$_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+X_2^4\alpha+Y_{-1}^0e$

Thus for mass number : 235 = 207+4X

4X= 28

X = 7

Thus for atomic number : 92 = 82+2X-Y

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14-10 = Y

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3 years ago

What are 2 applications of the Colligative Property that allows freezing point depression?

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3 years ago

Un anillo, de masa 90 gramos, contiene 59,1% de oro. ¿Cuál es el valor del anillo si cada mol de oro vale S/.1800?. Considere el

LekaFEV [45]

Answer:

S/.486 es el valor del anillo

Explanation:

Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.

Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:

90g × 59.1% = 53.19g Oro en el anillo

Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:

53.19g × (1mol / 197g) = <em><u>0.27 moles de oro contiene el anillo</u></em>.

Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:

0.27mol × (S/.1800 / 1mol oro) =

<h3>S/.486 es el valor del anillo</h3>

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3 years ago