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3 years ago

How do enzymes speed un chemical reactions?

1 answer:

8 0

Enzymes are extremely selective catalysts that help to speed up reactions. The molecules an enzyme works with are called substrates and they bind to the enzyme. This helps speed up reactions greatly once the substrates bind to the active site of the enzyme.

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in an experiment 3.5g of element A reacted with 4.0g of element G to form a compound Calculate the empirical formula for this co

kolezko [41]

Additional information

Relative atomic mass(Ar) : A=7, G=16

The empirical formula : A₂G

<h3>Further explanation</h3>

Given

3.5g of element A

4.0g of element G

Required

the empirical formula for this compound

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.

The empirical formula also shows the simplest mole ratio of the constituent elements of the compound

mol of element A :

$\tt mol=\dfrac{mass}{Ar}\\\\mol=\dfrac{3.5}{7}=0.5$

mol of element G :

$\tt mol=\dfrac{4}{16}=0.25$

mol ratio A : G = 0.5 : 0.25 = 2 : 1

4 0

3 years ago

39 which statement about the process of dissolving is TRUE

xz_007 [3.2K]

C !!!!!!!!!!!!!!!!!!!!!!

6 0

3 years ago

A little boy has placed his pet squirrel into a balloon with a volume of 3L at a pressure of 1.0 atm. If the boy takes his pet s

s2008m [1.1K]

Answer: The squirrel's balloon will be 0.86 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=1.0atm\\V_1=3L\\P_2=3.5atm\\V_2=?$

Putting values in above equation, we get:

$1.0\times 3L=3.5\times V_2\\\\V_2=0.86$

Thus the squirrel's balloon will be 0.86 L

7 0

3 years ago

Provided is a diagram showing forces on a box.

Eduardwww [97]

Answer:

5 N left

Explanation:

trust me cccuuhh

4 0

3 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 192.mg of oxalic acid H2

kap26 [50]

Answer:

The molarity of the NaOH solution is 0.076 M

Explanation:

Step 1: Data given

Mass of oxalic acid = 192 mg = 0.192 grams

volume = 250 mL = 0.250 L

Molar mass oxalic acid = 90.03 g/mol

Step 2: The balanced equation

H2C2O4 + 2NaOH → Na2C2O4 + H2O

Step 3: Calculate moles of oxalic acid

Moles oxalic acid = 0.192 grams / 90.03 g/mol

Moles oxalic acid = 0.00213 moles

Step 4: Calculate molarity of oxalic acid

Molarity = Moles / volume

Molarity = 0.00213 moles / 0.250 L

Molarity = 0.00852 M

Step 5: Calculate Molarity of NaOH

2 Ca*Va = Cb*Vb

with Ca = Molarity of oxalic acid = 0.00852 M

with Va = volume of oxalic acid = 0.250 L

with Cb = Molarity of NaOH = TO BE DETERMINED

with Vb = volume of 0.0558 L

Cb = (2*0.00852 * 0.25) / 0.0558

Cb = 0.076 M

The molarity of the NaOH solution is 0.076 M

4 0

3 years ago