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2 years ago

What is the test used to diffferentate Be(OH) from Ba(OH)?

1 answer:

6 0

Ba (OH) 2 Ba(OH) 2 is a string electrolyte, determine the concentration of each of the individual in a 0.400m Ba (OH)20.400m

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What happens when

jolli1 [7]

Answer:

a i think but idk

Explanation:

3 0

3 years ago

A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2

Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

$H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O$ (Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL$

Putting values in above equation, we get:

$2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL$

$M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M$

0.823 M was the molarity of the KOH solution.

7 0

3 years ago

Inorganic compounds which are found in the earth are called

saul85 [17]

Answer: The answer would be minerals. Synthetics are made by men, and shouldn't be found in the earth Vitamins are organic Cement is a mixture of an organic compounds and rarely found in nature.

6 0

3 years ago

The lock-and-key model and the induced-fit model are two models of enzyme action explaining both the specificity and the catalyt

ivolga24 [154]

Answer:

The lock-and-key model:

c. Enzyme active site has a rigid structure complementary

The induced-fit model:

a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.

Common to both The lock-and-key model and The induced-fit model:

b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.

d. Substrate binds to the enzyme through non-covalent interactions

Explanation:

Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.

The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.

The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.

4 0

3 years ago

30. The density of an unknown gas at 27°C and 2 atm pressure is equal with density of N2 gas at

Zanzabum

Answer:

Molar mass of the unknown gas is 64.6 g/mol

Explanation:

Let's think this excersise with the Ideal Gases Law.

We start from the N₂. At STP conditions we know that 1 mol of anything occupies 22.4L.

We apply: P . V = n . R . T

5 atm . V = 1 mol . 0.082 . 325K

V = (1 mol . 0.082 . 325K) / 5 atm = 5.33 L

It is reasonable to say that, if we have more pressure, we may have less volume.

As this is the volume for 1 mol of N₂, our mass is 28 g. Then, the density of the nitrogen and the unknown gas is 28 g/5.33L = 5.25 g/L

Our unknown gas has, this density at 27°C and 2 atm.

If we star from this, again: 1 mol of any gas occupy 22.4L at STP, we can calculate the volume for 1 mol at those conditions:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

1 atm . 22,4L / 273K = 2 atm . V₂ / 300K

Remember that the value for T° is Absolute (T°C + 273)

[ (1 atm . 22.4L / 273K) . 300K] / 2 atm = V₂ → 12.3L

This is the volume for 1 mol of the unknown gas at 2 atm and 27°C

We use density to determine the mass: 12.3 L . 5.25 g/L = 64.6 g

That's the molar mass: 64.6 g/mol

6 0

2 years ago