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marysya [2.9K]
3 years ago
15

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

Chemistry
2 answers:
nasty-shy [4]3 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]3 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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If Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?
Fiesta28 [93]
The given sentence is part of a longer question.

I found this question with the same sentence. So, I will help you using this question:

For the reaction N2O4<span>(g) ⇄ 2NO</span>2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm<span>).  If </span>Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?


(a) Q = K<span>;   The reaction </span>is at equilibrium.
(b) Q < K<span>;   The reaction </span>will proceed to the right.
(c) Q > K<span>;   The reaction </span>will proceed to the left.

The answer is the option (c) Q > K<span>; The reaction will proceed to the </span>left, since Qp<span> = </span>1<span>, and 1 > 0.15.</span>
Explanation:

Kp is the equilibrium constant in term of the partial pressures of the gases.

Q is the reaction quotient. It is a measure of the progress of a chemical reaction.

The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.

At equilibrium both Kp and Q are equal. Q = Kp

If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,

If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.

Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.

Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.
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3 years ago
What is the % composition of Carbon in Chromium (iii) Carbonate
photoshop1234 [79]

Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

\begin{gathered} C\rightarrow3\times12=36 \\  \\ O\rightarrow9\times16=144 \\  \\ Cr\rightarrow2\times52=104 \end{gathered}

The molar mass will be thus:

M=36+104+144=284\text{ g/mol}

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\  \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}

The percent composition of Carbon is thus 12.7 %.

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3 years ago
(06.03 MC) A 50.0 mL sample of gas at 20.0 atm of pressure is compressed to 40.0 atm of pressure at constant temperature. What i
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Answer:

New volume is 25.0 mL

Explanation:

Let's assume the gas sample behaves ideally.

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V_{1} and V_{2} represent initial and final volume respectively

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Here, T_{1}=T_{2}, V_{1}=50.0mL, P_{1}=20.0atm and P_{2}=40.0atm

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