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creativ13 [48]
3 years ago
5

Balancing Practice (some of these may be in the simulation)

Chemistry
1 answer:
agasfer [191]3 years ago
8 0

Answer:

see below

Explanation:

16. 1 P₄(s) + 6 F₂(g) → 1 PF₃(s)

17. 2 C(s) + 2 H₂O(g) → 1 CH₄(g) + 1 CO₂(g)

18. 2 HgO(s)→ 1 O₂(g) + 2 Hg(l)

19. 1 CaCO₃(s) → 1 CO₂(g) + 1 CaO(s)

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the rate of disappearance of Br- at some moment in time was determined to be 3.5 x 10-4 M/s. What is the rate of appearance of B
ddd [48]

Answer:

1.8 × 10⁻⁴ mol M/s

Explanation:

Step 1: Write the balanced reaction

2 Br⁻ ⇒ Br₂

Step 2: Establish the appropriate molar ratio

The molar ratio of Br⁻ to Br₂ is 2:1.

Step 3: Calculate the rate of appearance of Br₂

The rate of disappearance of Br⁻ at some moment in time was determined to be 3.5 × 10⁻⁴ M/s. The rate of appearance of Br₂ is:

3.5 × 10⁻⁴ mol Br⁻/L.s × (1 mol Br₂/2 mol Br⁻) = 1.8 × 10⁻⁴ mol Br₂/L.s

3 0
3 years ago
I2 is produced by the reaction of 0.4235 mol of CuCl2 according to the following equation: 2CuCl2+4KI⟶2CuI+4KCl+I2. What volume
sveta [45]

Answer: 4.7432 L

Explanation:

Use stoichiometry: .4235 mol CuCl2 (1 mol I2 / 2 mol CuCl2)(22.4 L / 1 mol I2) = 4.7432 L :)

6 0
2 years ago
Why might a scientist repeat an experiment if he/she did not make a mistake in the first one? An experiment should be repeated t
makvit [3.9K]

The answer is C. To ensure the results are accurate.

5 0
3 years ago
Read 2 more answers
A metallurgist has one alloy containing 21% aluminum and another containing 42% aluminum. how many pounds of each alloy must he
Naily [24]
Answer is: 7.8 lb of 21% aluminum and 33.2 ib of <span>42% aluminum.</span>

ω₁<span> = 21% ÷ 100% = 0.21.
ω</span>₂<span> = 42% ÷ 100% = 0.42.
ω</span>₃<span> = 38% ÷ 100% = 0.38.
</span>m₁ = ?.

m₂<span> = ?.
</span>m₃ = m₁ + m₂<span>.
</span>m₃ = 41 pounds.

m₁ = 41 lb - m₂<span>.
ω</span>₁ · m₁ + ω₂ ·m₂ = ω₃ · m₃.

0.21 · (41 lb - m₂) + 0.42 · m₂ = 0.38 · 41 lb.

8.61 lb - 0.21m₂ + 0.42m₂ = 15.58 lb.

0.21m₂ = 6.97 lb.

m₂ = 6.97 lb ÷ 0.21.

m₂ = 33.2 lb.

m₁ = 41 lb - 33.2 lb.

m₁<span> = 7.8 lb.</span>



8 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
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