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2 years ago

Why is it better to use the metric system, rather than the English system, in scientific measurement?

Chemistry

2 answers:

nasty-shy [4]2 years ago

6 0

Answer:

Explanation:

Hello,

English system is indeed based on the metric system because they look for reference stuff to get the measurements, for instance, a foot equals 30.48 cm, but a common 30-cm rule measures every 1 cm, so the metric system also provides more accurate measurements. An other example lies on the measurement of mass, a pound which equals 453.59 g is appropriated to measure middle amounts of masses, but for example, in qualitative analysis we tend to measure amounts below about 0.10 g (0.022 lb) so it is easier for us in science to use the metric system in addition to the straightforwardness of the reporting data in 10-based notation (scientific notation).

Best regards.

frez [133]2 years ago

3 0

The metric system is based on powers of 10 so it is much easier to convert units, often just by moving the decimal point.

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A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain

tatiyna

Answer:

$C_m=0.474\frac{J}{g\°C}$

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

$Q_m=-Q_w$

Thus, in terms of masses, specific heats and temperatures we can write:

$m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)$

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

$C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}$

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

$C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}$

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