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Wittaler [7]
3 years ago
11

Which plastic do you think could best contain extremely corrosive material? PETE, LDPE, PVC, PS, PP, or HDPE.

Chemistry
1 answer:
Viktor [21]3 years ago
7 0
The ones that are known to be corrosive are PETE and PVC. These plastics you need to avoid because these are <span>plastics that can be hydrolyzed by alkaline solutions. Hope this helps on your task</span>
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A typical aspirin tablet contains 5.00 grains of pure aspirin analgesic compound. The rest of the tablet is
Pavel [41]

Answer:

Tablets=193

Explanation:

Hello,

In this case, given that a typical aspirin tablet contains 5.00 grains of pure aspirin, the first step here is to compute the mass of those grans per tablet given that 1.00 g = 15.4 grains:

m=5.00grains*\frac{1.00g}{15.4grains}=0.325g

In such a way, the number of aspirin tablets are computed considering the total mass of aspirin and the mass per tablet:

Tablets=\frac{62.7g}{0.325g}\\ \\Tablets=193

Best regards.

8 0
3 years ago
H2(g)+p4(s)ph3(g) what the answer ?
disa [49]

Answer:

6H2 + P4→ 4PH3

Explanation:

Phosphorus has 4 in it and hydrogen has 3 in it. in order to balance it, we have to put 4 in front of phosphine so that the phosphorus on the product side has an equal amount as to the one on the reactant side.

the only one left to balance is hydrogen and so in order to balance it we put a 6 on h2 because the hydrogen in the product size becomes 12 (4 * 3).

therefore the hydrogen on the reactant side becomes 12 as well (6 * 2)

8 0
3 years ago
How many moles of carbon dioxide are produced from 8.73 moles of tristearin?​
matrenka [14]

Answer:

Explanation:

  8.73 mol                                         x mol

2 C57H110O6(S) + 163 O2(g)  --->  114 CO2(g) + 110 H2O(I)

2 mol                                               114 mol

8.78 mol (114mol/2 mol) =500.46 mol

8 0
3 years ago
A compound contains only change and n combustion of 35.0mg of the compound produces 33.5mg co2 and 41.1mg h2o. What is the empir
Viefleur [7K]

Answer:

The empirical formula is CH6N2

Explanation:

A compound containing only C, H, and N yields the following data. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and 41.1 mg of H2O. What is the empirical formula of the compound

Step 1: Data given

Mass of the compound = 35.0 mg = 0.035 grams

Mass of CO2 = 33.5 mg = 0.0335 grams

Mass of H2O = 41.1 mg = 0.0411 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 0.0335 grams / 44.01 g/mol

Moles CO2 = 7.61 *10^-4 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 7.61 *10^-4 moles CO2 we have 7.61 *10^-4 moles C

Step 4: Calculate mass C

Mass C = 7.61 *10^-4 moles * 12.01 g/mol

Mass C = 0.00914 grams = 9.14 mg

Step 5: Calculate moles H2O

Moles H2O = 0.0411 grams / 18.02 g/mol

Moles H2O = 0.00228 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.00228 moles H2O we have 2* 0.00228 = 0.00456 moles H

Step 7: Calculate mass H

Mass H = 0.00456 moles * 1.01 g/mol

Mass H = 0.00461 grams = 4.61 mg

Step 8: Calculate mass N

Mass N = 35.0 mg - 9.14 - 4.61 = 21.25 mg = 0.02125 grams

Step 9: Calculate moles N

Moles N = 0.02125 grams / 14.0 g/mol

Moles N = 0.00152 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moes

C: 0.000761 moles / 0.000761 moles= 1

H:  0.00456 moles / 0.000761 moles = 6

N: 0.00152 moles  / 0.000761 moles = 2

For every C atom we have 6 H atoms and 2 N atoms

The empirical formula is CH6N2

5 0
3 years ago
after a chemistry student made a AgNO3(small 3 at bottom) solution., she wanted to determine the molar concentration of it. if 2
Alexxandr [17]
Molar concentration = (numbet of mol Solute)/ ( volume Solution)

1) Finding
 the number of the mol solute

22.0 g AgNO3 * \frac{1 mol AgNO3}{169.91 g AgNO3} = 0.129 mol AgNO3&#10;&#10;2) 725 ml = 0.725 L&#10;&#10;&#10;

3) Molarity = \frac{number/ mol/solute}{Volume/solution} = \frac{0.129}{0.725} &#10;&#10;

Molarity= \frac{0.178 mol}{L} , &#10;&#10;or 0.178 M&#10;


3 0
3 years ago
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