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3 years ago

5

How can you force a solid compound to the bottom of a melting point capillary? a. tap the closed endb. tap the open endc. use a

needle

1 answer:

snow_lady [41]3 years ago

3 0

Answer: option A. tap the closed end

Explanation:

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Solve the quadratic equation 2x^2+13x=15 by method of completing the square<br>

mafiozo [28]

Answer:

x = 1, -7.5

Explanation:

2x² + 13x = 15

Divide the left side of the equation by 2

2(x² + 6.5x) = 15

Divide 6.5 by 2 and square the quotient

6.5/2 = 3.25

3.25² = 10.5625

Add 10.5625 to the left side

2(x² + 6.5x + 10.5625) = 15

Since you have a 2 outside the parentheses, you will be adding 10.5625•2 to the right side.

10.5625 • 2 = 21.125

2(x² + 6.5x + 10.5625) = 36.125

To factor (x² + 6.5x + 10.5625), add b/2 to x

b/2 = 6.5/2 = 3.25

2(x + 3.25)² = 36.125

Divide by 2

(x + 3.25)² = 18.0625

Square root.

(x + 3.25) = √18.0625

x + 3.25 = ±4.25

Subtract 3.25.

x = 4.25 - 3.25 = 1

x = -4.25 - 3.25 = -7.5

x = 1, -7.5

4 0

3 years ago

One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu

Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

7 0

3 years ago

Which of the following is most likely to be found in a crystalline structure

aev [14]

The correct answer is letter D: Quartz.

5 0

4 years ago

To balance a chemical equation it may be necessary to adjust the

Mars2501 [29]

Answer:

<em>To balance a chemical equation it may be necessary to adjust the </em><u>coefficients.</u>

Explanation:

The <em>coefficients</em> of a <em>chemical equation</em> are the numbers that you put in front of each reactant and product. They are used to balance the equation and comply with the law of mass conservation.

By adjusting the coefficients you obtain the relative amounts (moles) of each product and reactant, i.e. the mole ratios.

Here an example.

The first information is what is called a word equation. E.g. nitrogen and hydrogen react to form ammonia:

Word equation: hydrogen + nitrogen → ammonia

Skeleton equation: H₂ + N₂ → NH₃

This equation shows the chemical formulae but it is not balanced. The law of mass conservation is not observed.

So, in order to comply with the law of mass conservation you adjust the coefficients as follow.

Balanced chemical equation: 3H₂ + N₂ → 2NH₃

As you see, it was necessary to modify the coefficients. Now the law of conservation of mass is observed and you get the mole ratios:

3 mol H₂ : 1 mol N₂ : 2 mol NH₃

4 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago