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3 years ago

Ty for anyone helping!

Chemistry

1 answer:

sp2606 [1]3 years ago

6 0

23. Enzyme A, as the temperature increases, the rate of reaction goes down. However, Enzyme B as the temperature goes up the reaction goes up.

24. Enzyme B is more active in a human cell, the reaction is what you want and everytime the temperature goes up in enzyme B the reaction goes up.

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A plane flew for 2 hours at 467 mph (mile per hour)and 5 hours at 536 mph. How far did the plane fly miles?

Andreyy89

Answer:

the plane flew 3614 miles

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3 years ago

How much water would be needed to completely dissolve 1.52 L of the gas at a pressure of 730 torr and a temperature of 21 ∘C?

Sophie [7]

Answer:

The correct answer is 0.4 L.

Explanation:

The mentioned question can be solved by using the equation,

C = K × Pgas--------(i)

Here K is the Henry law constant whose value is 0.158 mol/L/atm, C is the concentration of the gas in liquid state, and Pgas is the partial pressure of the gas.

Now to find the volume of water, the formula to be used is,

PV = nRT-----------(ii)

Here P is the pressure of the gas, V is the volume, R is the universal gas constant whose value is 0.082 Latm/mol/K and T is the temperature.

PgasV = nRT

Pgas = nRT/Vgas

The value of Pgas is inserted in equation (i) we get,

C = K × nRT/Vgas

It is to be noted that C = n/V, here n is the no. of the moles and V is the volume of liquid.

n/Vliquid = K × nRT/Vgas

1/Vliquid = KRT/Vgas

Vliquid = Vgas/KRT--------------(iii)

Based on the given information, the volume of the gas is 1.52 L, the value of K is 0.158 mol/L/atm, the value of R is 0.082 Latm/mol/K and value of T is 21 degree C or 273 + 21 = 294 K.

Now putting the values in equation (iii) we get,

Vliquid = 1.52 L / 0.158 × 0.082 × 294

Vliquid = 1.52 / 3.809

Vliquid = 0.399 or 0.4 L

Hence, the volume of water required to dissolve 1.52 L of gas is 0.4 L.

8 0

3 years ago

1. Consider the decomposition reaction of sodium chlorate. There are 100 grams of

IRINA_888 [86]

A. NaCl(s) and O2(g)

B. 2NaClO3(s) —> 2NaCl(s) + 3O2(g)

C. moles NaClO3 = 100 g / 106.44 g/mol = 0.939 mol NaClO3

D. 0.939 mol NaCl (because the NaClO3 and NaCl are in a 1 to 1 ratio)

E. grams NaCl = 0.939 mol • 58.44 g/mol = 54.9 g NaCl

F. moles of O2 = 0.939 mol NaClO3 • (3 mol O2 / 2 mol NaClO3) = 1.41 mol O2

G. grams of O2 = 1.41 mol • 32 g/mol = 45.1 g O2

H. Percent yield = 10/45.1 • 100% = 22.2% yield

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3 years ago

Describe El Niño and La Niña in your own words.

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2 years ago