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1 year ago

The equations ? It’s for a chemistry class, and he said we need to explain and write down the math problems?

Chemistry

1 answer:

alina1380 [7]1 year ago

3 0

Answer:

I think this is because math and chemistry go together and the math problems are science related.

Explanation:

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What is the purpose of the uninoculated control tubes used in the oxidation fermentation test?

Black_prince [1.1K]

The purpose of the uninoculated control tubes used in this test is that two uninoculated control tubes are needed to show the results of the medium in both aerobic and anaerobic environments. It is used to show it is sterile and also as a color comparison, used also to show that the medium remains green under both conditions.

5 0

2 years ago

What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen

photoshop1234 [79]

Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

4Al + 3O2 —> 2Al2O3

From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol

Mass of Al2O3 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

8 0

2 years ago

Complete the sentence.

Leno4ka [110]

Answer:

6, double

Explanation:

Hex- is a prefix for number 6.

Ene- is a suffix for a double bond.

7 0

2 years ago

A rigid container is filled with chlorine gas. The gas has a pressure of 2.75 bar. The tank is then cooled down to -20.0oC at wh

Gnom [1K]

Answer:

Original temperature (T1) = - 37.16°C

Explanation:

Given:

Gas pressure (P1) = 2.75 bar

Temperature (T2) = - 20°C

Gas pressure (P2) = 1.48 bar

Find:

Original temperature (T1)

Computation:

Using Gay-Lussac's Law

⇒ P1 / T1 = P2 / T2

⇒ 2.75 / T1 = 1.48 / (-20)

⇒ T1 = (2.75)(-20) / 1.48

⇒ T1 = -55 / 1.48

⇒ T1 = - 37.16°C

Original temperature (T1) = - 37.16°C

3 0

2 years ago

A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t

andrezito [222]

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

= 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

8 0

2 years ago