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Vilka [71]
3 years ago
13

Rank the following quantities from smallest to largest: 5,200, dozen, pair,gross,ream.

Chemistry
1 answer:
Harrizon [31]3 years ago
6 0

5,200


dozen = 12


pair = 2


gross = 144


ream =  500


Pair, dozen, gross, ream, 5,200

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ira [324]

Answer:

1) 40

2) 2.25 moles

3) 17

4) 120

5) Fe₂O₃

Explanation:

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6 0
3 years ago
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3S8 + 8 OH- + 8 S3 + 4 HOOH
NemiM [27]

Answer:

Second order

Explanation:

We could obtain the order of reaction by looking at the table very closely.

Now notice that in experiment 1 and 2, the concentration of [OH^-] was held constant while the concentration of [S8] was varied.  So we have;

a situation in which the rate of reaction was tripled;

0.3/0.1 = 2.10/0.699

3^1 = 3^1

Therefore the order of reaction with respect to  [S8] is 1.

For [OH^-], we have to look at experiment 2 and 3 where the concentration of [S8] was held constant;

x/0.01 = 4.19/2.10

x/0.01 = 2

x = 2 * 0.01

x = 0.02

So we have;

0.02/0.01 = 2^1

2^1 = 2^1

The order of reaction with respect to  [OH^-] = 1

So we have the overall rate law as;

Rate = k[S8]^1  [OH^-] ^1

Overall order of reaction = 1 + 1 = 2

Therefore the reaction is second order.

4 0
3 years ago
Which of these is an example of food engineering?
Andrej [43]

Answer:

a.labeling food with an ingredient list to inform consumers

Explanation:

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3 years ago
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What is the formula for germicide that is sufficient to kill blood-borne pathogens?
Colt1911 [192]
It is 1 ounce. Bloodborne Pathogens can be transmitted when blood or body liquid from a tainted individual enters someone else's body by means of needle-sticks, human chomps, cuts, scraped areas, or through mucous films. Likewise, semen, vaginal discharges and salivation in dental methods are considered conceivably tainted body liquids.
7 0
3 years ago
Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon
Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: Cu_2O\text{ and }CuO

  • <u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = \frac{2.26}{2.26}=1

For Oxygen  = \frac{4.55}{2.26}=2.01\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is CO_2

  • <u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = \frac{3.57}{3.57}=1

For Oxygen  = \frac{3.57}{3.57}=1

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is CO

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = \frac{1}{2}\times 1=\frac{1}{2}

Now, taking the ratio of carbon atoms in both the samples, we get:

C_1:C_2=\frac{1}{2}:1=1:2

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0
3 years ago
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