What is the relationship between plastic and gasoline?

tekilochka [14]

Answer:

Sr would be the limiting reactant

5 moles

Explanation:

Since the equation is a balanced equation, the coefficient shows how each substance relates to the other in terms of the number of moles.

Reactants would be those on the left hand side of the arrow, while the products would be found on te right and side of the arrow. In this question, the reactants would be Sr and O₂.

Limiting reactant is the reactant that is insufficient; meaning to say that there is not enough of that substance and thus the reaction cannot continue. The other reactant(s) that is not limiting is called the excess reactants.

From the balanced equation, 2 moles of Sr is needed to react with 1 mole of O₂. Thus, if we have 5 moles of each reactant, Sr would be the limiting reactant since for every 1 mole of O₂, there has to be 2 moles of Sr in order for the reaction to proceed. Thus, if we have 5 moles of O₂, we would need 10 moles of Sr.

When we work out the amount of products formed, we look at the number of moles of the limiting reactant. This is because the limiting reactant determines how much is being reacted, while the excess number of moles of the excess reactant will remain unreacted.

For every 2 moles of Sr reacted, 2 moles of SrO would be produced. This means that the mole ratio of Sr to SrO is 1:1. Thus, since 5 moles of Sr has been reacted, 5 moles of the product (SrO) would be produced.

8 0

2 years ago

Phenolphthalein indicator is??

Aleonysh [2.5K]

Answer:

Hey mate.....

Explanation:

This is ur answer.....

Phenolphthalein is often used as an indicator in acid–base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions. It belongs to the class of dyes known as phthalein dyes.

Hope it helps!

mark me brainliest pls....

Follow me! :)

7 0

2 years ago

A 0.5895-g sample of impure magnesium hydroxide is dissolved in 100.0 mL of 0.2050 M HCl solution. The excess acid then needs 19

Zepler [3.9K]

Answer:

91.38 %

Explanation:

First we use the following formula to calculate the number of moles of HCl and NaOH contained in the solutions:

molar concentration = number of moles / solution volume

number of moles = molar concentration × solution volume

number of moles of HCl = 0.2050 × 100 = 20.5 mmoles

number of moles of NaOH = 0.1020 × 19.85 = 2.02 mmoles

reaction between NaOH and HCl:

NaOH + HCl → NaCl + H₂O

If 2.02 mmoles of NaOH were used for neutralization of HCl left from the reaction, which are 2.02 mmoles of HCl, it means that the HCl reacted with Mg(OH)₂:

20.5 - 2.02 = 18.48 mmoles of HCl

2 HCl + Mg(OH)₂ = MgCl₂ + 2 H₂O

From the chemical reaction we devise the following reasoning:

if 2 mmoles of HCl react with 1 mmole of Mg(OH)₂

then 18.48 mmoles HCl react with X mmoles of Mg(OH)₂

X = (18.48 × 1) / 2 = 9.24 mmoles of Mg(OH)₂

now the mass of Mg(OH)₂:

mass = number of moles × molecular weight

mass of Mg(OH)₂ = 9.24 × 58.3 = 538.7 mg = 0.5387 g

Now the percent by mass of magnesium hydroxide Mg(OH)₂ in the impure sample (purity):

purity = (mass of pure Mg(OH)₂ / mass of impure sample) × 100

purity = (0.5387 / 0.5895) × 100

purity = 91.38 %

7 0

2 years ago

HELP ME PLEASE PLEASE HELP TAHNK YOU :)

allsm [11]

If a light bounced off at a 40 degree angle it would bounce off at an 80 degree angle. Hope i am right

4 0

2 years ago

Glucose (C6H12O6) is a key nutrient for generating chemical potential energy in biological systems. We were provided 16.55 g of

harina [27]

Answer:

a) 40 %

b) $4.04~g~CO_2$

c) $5.53x10^2^3~molecules~of~O_2$

Explanation:

For a) we will have to calculate the molar mass of $C_6H_1_2O_6$ , so the first step is to find the atomic mass of each atom and multiply by the amount of atoms in the molecule.

C => 12*(6) = 72

H => 1*(12) = 12

O => 6*(16) = 96

Molar mass = 180 g/mol

Then we can calculate the percentage by mass:

$Percentage~=~\frac{72}{180}*100=40$

For b) we have to start with the reaction of glucose:

$C_6H_1_2O_6~+~6O_2~->~6CO_2~+~6H_2O$

Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the following conversion ratios:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol carbon dioxide

-) 1 mol carbon dioxide = 44 g carbon dioxide

$16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~CO_2}{1~mol~C_6H_1_2O_6}\frac{44~g~CO_2}{1~mol~CO_2}=4.04~g~CO_2$

For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the following conversion ratios:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol oxygen

-) 1 mol oxygen = 6.023x10^23 molecules of O2

$16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~O_2}{1~mol~C_6H_1_2O_6}\frac{6.023x10^2^3~molecules~O_2}{1~mol~O_2}=~5.53x10^2^3~molecules~of~O_2$

4 0

2 years ago