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3 years ago

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Chemistry

1 answer:

ipn [44]3 years ago

7 0

Answer:

Explanation:

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A welder has 1.873 × 102 g Fe2O3 and 94.51 g Al in his welding kit. Which reactant will he run out of first?

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_______are lipids that contain cholesterol

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Answer: Fats

Explanation: Cholesterol is one of several types of fats (lipids) that play an important role in your body.

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Match the following words.

azamat

Answer:

1. active site

2. substrate.

3. activity

4. cofactor

5. complex

6. coenzyme.

Explanation:

1. A substrate binds to an enzyme at the active site where the reaction occurs.

2. In a catalyzed reaction, a reactant is often called a substrate.

3 activity is a measure of how fast an enzyme can convert the reagent to the product.

4. An inorganic substance necessary for the function of some enzymes is called a cofactor.

5. When properly aligned, the enzyme and substrate form an enzyme-substrate (ES) complex.

6. A small organic molecule necessary for the function of some enzymes is called a coenzyme.

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Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

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What is the solubility of Na2HAsO4 at 60 degrees C

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C. 65g / 100ml of water

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