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Ksju [112]
3 years ago
15

Which term names enzymes in the body that speed up chemical reactions?

Chemistry
2 answers:
Diano4ka-milaya [45]3 years ago
8 0

Answer:

B: Catalysts

Explanation:

Viefleur [7K]3 years ago
7 0

Answer:

Catalyst speeds up the chemical reactions. Catalyst is the compound that adds to the chemical reaction so that it happen more quickly. Catalyst without being consumed in the reaction accelerates the reaction. It lowers down the activation energy of the transition state and thus accelerates the reaction.

Hope this helped!

Plz mark me as Brainliest!

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PLEASE HELPPP!!!!!!!!
Stells [14]

Explanation:

Label A: oceanic - oceanic

Label B: oceanic - continental

Label C: continental - continental

5 0
3 years ago
Read 2 more answers
He equation for the dissociation of pyridine is
sergeinik [125]

Answer:

10.10

Explanation:

Step 1: Write the basic dissociation reaction for pyridine

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq)      Kb = 1.9 × 10⁻⁹

Step 2: Calculate [OH⁻]

For a weak base, we will use the following expression.

[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M

Step 3: Calculate pOH

We will use the definition of pOH.

pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9

Step 4: Calculate pH

We will use the following expression.

pH = 14 - pOH = 14 - 3.9 = 10.10

3 0
3 years ago
The heat of fusion for water is 80. cal/g. How many calories of heat are needed to melt a 35 g ice cube that has a temperature o
HACTEHA [7]

Answer: 2800 calories

Explanation:

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.

Amount of heat required to fuse 1 gram of water = 80 cal

Mass of ice given = 35 gram

Heat required to fuse 1 g of ice at 0^0C = 80 cal

Thus Heat required to fuse 35 g of ice =\frac{80}{1}\times 35=2800cal

Thus 2800 calories of energy is required to melt 35 g ice cube

6 0
4 years ago
What is the percent yield of ferrous sulfide if the actual yield is 220.0g and the theoretic yield is 275.6?
Amanda [17]
ANSWER yield = 79.83%
3 0
3 years ago
Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W
nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

     H2= (1-x)Haq+XHvap.........1

    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

                        = 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

                     x = 1904.976 (J/g)/1998.576 (J/g)

                     x = 0.953

So, the quality of the wet steam is 0.953

                   

7 0
3 years ago
Read 2 more answers
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