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3 years ago

The intake stroke of a positive displacement compressor is most similar to that of a

Chemistry

1 answer:

umka21 [38]3 years ago

5 0

Answer:

D. two-stroke cylinder engine.

Explanation:

Two-stroke cylinder engine: In chemistry, the term "two-stroke cylinder engine" is described as one of the types of "internal combustion engine" that tends to complete a single "power cycle" possessing two different strokes of the "piston" during a particular "crankshaft revolution".

It generally consists of two different strokes that are being often known as power or exhaust and suction stroke.

In the question above, the given statement signifies the "two-stroke cylinder engine". 

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A solution contains one or more of the following ions: Ag , Ca2 , and Co2 . Lithium bromide is added to the solution and no prec

koban [17]

Answer:

$Ca^{2+} and Co^{2+}$

Explanation:

Given:

A solution contains one or more of the following ions such as Ag, $Ca_2$ and $Co_2$

Here the Lithium bromide is added to the solution and no precipitate forms

Solution:

Since with LiBr no precipitation takes place therefore Ag+ is absent

Here on adding $Li_2SO_4$ to it precipitation takes place.

Precipitate is as follows,

$Ca^{2+}(aq)+SO_4^2^-(aq)----->CASO_4(s)$

Thus,

$Ca^2^+$ is present

When $Li_3PO_4$ is added again precipitation takes place.

Therefore the reaction is as follows,

$Co^2^+(aq)+PO_4^3^-(aq)------>Co_3(PO_4)_2(s)$

Therefore,

$Ca^{2+} and Co^{2+}$ are present in the solution

3 0

3 years ago

What is the value of the activation energy of the uncatalyzed reaction?

Lubov Fominskaja [6]

The value of the activation energy of an uncatalyzed reaction is greater than that of a catalyzed reaction. As we know, a catalyst provides an alternative path for the reaction to happen at a faster rate. So, for a catalyzed reactio, activation energy is lesser than the original path.

6 0

3 years ago

It is required to prepare 1250 kg of a solution composed of 12 wt.% ethanol and 88 wt.% water. Two solutions are available, the

bekas [8.4K]

Answer:

437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.

Explanation:

Let the mass of the first solution be x and second solution be y.

Amount solution required = 1250 kg

x + y = 1250 kg....[1]

Percentage of ethanol in required solution = 12% of 1250 kg

Percentage of ethanol in solution-1 = 5% of x

Percentage of ethanol in required solution = 25% of y

5% of x + 25% of y =12% of 1250 kg

$\frac{5}{100}\times x+\frac{25}{100}y=\frac{12}{100}\times 1250 kg$

x + 5y = 3000 kg...[2]

Solving [1] and [2] we :

x = 437.5 kg , y = 812.5 kg

437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.

8 0

3 years ago

32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are plac

Lina20 [59]

Answer:

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

Explanation:

Step 1: Data given

Mass of sulfur = 32.00 grams

Mass of oxygen = 48.00 grams

Molar mass of sulfur = 32.07 g/mol

Molar mass of oxygen = 32 g/mol

Molar mass of SO3 = 80.07 g/mol

Step 2: The balanced equation

2S + 3O2 → 2SO3

Step 3: Calculate moles S

Moles S = Mass S / molar mass S

Moles S = 32.0 grams / 32.07 g/mol

Moles S = 0.998 moles

Step 4: Calculate moles O2

Moles O2 = 100.0 grams / 32.0 g/mol

Moles O2 = 3.125 moles

Step 5: Calculate the limiting reactant

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

S is the limiting reactant. It will completely be consumed (0.998 moles)

O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles

There will remain 3.125- 1.497 = 1.628 moles O2

This is 1.628 moles * 32 g/mol = 52.1 grams

Step 6: Calculate moles SO3

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

For 0.998 moles S there will react 0.998 moles SO3

Step 6: Calculate mass SO3

Mass SO3 = moles SO3 * molar mass SO3

Mass SO3 = 0.998 moles * 80.07 g/mol

Mass SO3 = 79.9 grams ≈ 80 grams

There will be produced 80 grams of SO3

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

4 0

3 years ago

If you could spend a billion dollars (1 x 10 to the 9th power) per second, how many years would it take to spend one mole of dol

lord [1]

1000 to the power 4 not so good at math but

3 0

3 years ago