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ANEK [815]
3 years ago
12

Does fat and soap have density? ​

Physics
2 answers:
irinina [24]3 years ago
5 0

Fat has a density of 0.92 g/cm3 and soap has a density of 0.84 g/cm3. These are all properties that make fat and soap different substances.

yawa3891 [41]3 years ago
3 0

Answer:

The density of fat is about .88 g/ml and the density of soap is about 1.18 g/ml.

Explanation:

Fat and soap are not the same substances one reason is they have different properties .Malleability, hardness, solubility, melting point, color, and density are all properties. A property is a characteristic of a substance that does not change.

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5g of sugar is dissolved in "50g" of water what is the mass of the sugar water
polet [3.4K]

55g because you would still have the weight of the sugar but it combines with the liquid and spreads out into the water.

4 0
3 years ago
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A block is 10cm long, 5cm wide and 2cm high and weighs 100g. What is the volume of the block? What is the density?
Ivahew [28]

Answer:

1gm/cm^3

Explanation:

its the answer

5 0
3 years ago
A 60cm long string of an ordinary guitar is tuned to produce the note a4 (frequency 440hz) when vibrating in its fundamental mod
pishuonlain [190]

Answer:

0.78 m

Explanation:

The relationship between wavelength and frequency of a wave is given by

v=f \lambda

where

v is the speed of the wave

f is the frequency

\lambda is the wavelength

For the sound wave in this problem, we have

f=440 Hz is the frequency

v = 344 m/s is the speed of sound in air

Substituting into the equation and re-arranging it, we find the wavelength:

\lambda=\frac{v}{f}=\frac{344 m/s}{440 Hz}=0.78 m

8 0
3 years ago
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Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

dS = \frac{\delta Q}{T} (1)

Donde:

Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

dS = \frac{L_{v}}{T}\cdot dm (1b)

Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0
3 years ago
Which of the
kompoz [17]
C. Amount of oxygen

The others either change but don’t decrease or they increase.
8 0
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