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3 years ago

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um bequer muito utilizado em laboratorio mede volumes de ate 250 ml.determine esse volume em: a)centimetros cubicos b)litros c)m

etros cubicos d)decimetros cubicos

1 answer:

Lelechka [254]3 years ago

6 0

Can u put it in english not be racist but i love chemistry i just dont know foreign language

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State two types of plastics and their uses.

Svetllana [295]

Thermoplastic and thermosetting

thermoplastic:- they are easily molded and extruded into films, fibres and packaging.For eg. PVC

thermosetting:-they are hard and durable and can be used for aircraft parts,tires and auto parts .For eg. phenolic resins.

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1 year ago

Read 2 more answers

CONCLUSIONS: Suppose a student titrated a sample of monoprotic acid of unknown concentration using a previously standardized sol

Zolol [24]

Concentration of unknown acid is 0.061 M

Given:

Concentration of NaOH = 0.125 M

Volume of NaOH = 24.68 mL

Volume of acid solution = 50.00 mL

To Find:

Concentration of the unknown acid

Solution: Concentration is the abundance of a constituent divided by the total volume of a mixture. The concentration of the solution tells you how much solute has been dissolved in the solvent

Here we will use the formula for concentration:

M1V1 = M2V2

0.125 x 24.68 = 50 x M2

M2 = 0.125 x 24.68 / 50

M2 = 0.061 M

Hence, the concentration of unknown acid is 0.061 M

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1 year ago

Identify the correct coefficients to balance the redox reaction with the lowest possible integer coefficients.

Monica [59]

Answer:

$\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag$ .

Explanation:

Electrons are conserved in a chemical equation.

The superscript of $\rm Ag^{1+}$ indicates that each of these ions carries a charge of $+1$ . That corresponds to the shortage of one electron for each $\rm Ag^{+}$ ion.

Similarly, the superscript $+3$ on each $\rm Al^{3+}$ ion indicates a shortage of three electrons per such ion.

Assume that the coefficient of $\rm Ag^{+}$ (among the reactants) is $x$ , and that the coefficient of $\rm Al^{3+}$ (among the reactants) is $y$ .

$\rm \mathnormal{x}\; Ag^{1+} + ?\; Al \to \mathnormal{y}\; Al^{3+} + ?\; Ag$ .

There would thus be $x$ silver ( $\rm Ag$ ) atoms and $y$ aluminum ( $\rm Al$ ) atoms on either side of the equation. Hence, the coefficient for $\rm Al\!$ and $\rm Ag\!$ would be $y\!$ and $x\!$ , respectively.

$\rm \mathnormal{x}\; Ag^{1+} + \mathnormal{y}\; Al \to \mathnormal{y}\; Al^{3+} + \mathnormal{x}\; Ag$ .

The $x$ $\rm Ag^{1+}$ ions on the left-hand side of the equation would correspond to the shortage of $x$ electrons. On the other hand, the $y$ $Al^{3+}$ ions on the right-hand side of this equation would correspond to the shortage of $3\, y$ electrons.

Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of $x$ electrons, the right-hand side should also be $x\!$ electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of $3\, y$ electrons. These two expressions should have the same value. Therefore, $x = 3\, y$ .

The smallest integer $x$ and $y$ that could satisfy this relation are $x = 3$ and $y = 1$ . The equation becomes:

$\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag$ .

6 0

2 years ago

High-pressure liquid chromatography (HPLC) is a method used in chemistry and biochemistry to purify chemical substances. The pre

kotykmax [81]

Answer:

3.41 x10⁶ torr

Explanation:

To solve this problem we need to remember the equivalency:

1 torr = 133.322 Pa

Then we can proceed to<u> convert 4.55×10⁸ Pa into torr.</u> To do that we just need to multiply that value by a fraction number, putting the unit that we want to convert <em>from</em> in the <em>denominator</em>, and the value we want to convert <em>to</em> in the <em>numerator</em>:

4.55x10⁸ Pa * $\frac{1torr}{133.322Pa} =$ 3.41 x10⁶ torr

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3 years ago

Neptunium-237 was the first isotope of a transuranium element to be discovered. the decay constant is /s. what is the half-life

kari74 [83]

The half-life in years of Neptunium-237 which was the first isotope is 2.1 $10^{6}$ years.

Neptunium is most stable and Neptunium-237 is undergoes alpha decay, it means Neptunium-237 is decays by the emission of alpha particles . Seven alpha particles is emitted during decay of Neptunium-237. Neptunium-237 is radioactive actinide elements and first transuranium element.

The transuranium synthesis process involves creating a transuranium element through the transmutation process . The transmutation process is the process of creating heavy elements from light elements. Hence the process is the transmutation of light elements. There are two types: artificial and natural transmutation.

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1 year ago