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2 years ago

C12H26O + SO3+NaOH ----> C12H25NaSO4+ H2O

Chemistry

1 answer:

Vlad [161]2 years ago

4 0

Answer:

We can solve this by the method of which i solved your one question earlier

so again here molar mass of C12H25NaSO4 is 288.372 and number of moles for 11900 gm of C12H25NaSO4 will be = 11900/288.372

which is almost = 41.26 moles

so to get one mole of C12H25NaSO4 we need one mole of C12H26O

so for 41.26 moles of C12H25NaSO4 it will require 41 26 moles of C12H26O

so the mass of C12H26O = 41.26× its molar mass

C12H26O = 41.26×186.34

= 7688.38 gm!!

so the conclusion is If you need 11900 g of C12H25NaSO4 (Sodium Lauryl Sulfate) you need C12H26O 7688.38 gm !!

Again i d k wether it's right or wrong but i tried my best hope it helped you!!

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One lap of an engineered hamster track measures 255 cm. To run 10. meters, how many laps should the hamster run? Report your ans

Maru [420]

Answer:

3.9 laps

Explanation:

Step 1: Given data

Length of the hamster track (L): 255 cm

Total distance to be run (D): 10 m

Step 2: Convert "D" to centimeters

We will use the relationship 1 m = 100 cm.

10 m × (100 cm/1 m) = 1.0 × 10³ cm

Step 3: Calculate the number of laps (n) that the hamster should run

We will use the following expression.

n = D/L

n = 1.0 × 10³ cm/255 cm

n = 3.9

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3 years ago

3.) this graph shows the rates of reaction in a chemical reaction with and without the addition of an enzyme

mars1129 [50]

Answer:

a. Concave down

Linear increasing

b. Increases the reaction rate

c. The reaction approaches the saturation point of the enzyme

Explanation:

a. For the reaction with enzyme, the shape is concave down. The action of the enzyme on the preferred substrate is initially very rapid and decreases as the enzyme becomes saturated and the ratio of products to substrate increases to approach an equilibrium rate of reaction

For the reaction without enzyme, the shape is linear and increasing. Increase in the concentration of the substrate will increase the number of effective collisions that lead into product formation leading to an increased rate of the chemical reaction

b. The enzyme increases the proportion of effective combination of substrates to form the products

c. The curve of the reaction with enzyme flattens out because as the concentration of the substrate increases while that of the enzyme remains the same, the enzyme becomes saturated and less able to increase the rate of the reaction of the excess substrate.

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2 years ago

3. What keeps Earth revolving around the sun?

DENIUS [597]

Answer:

A. The force of gravity.

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2 years ago

Read 2 more answers

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago

A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal

bixtya [17]

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V = Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

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2 years ago