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MA_775_DIABLO [31]
3 years ago
7

An increase in enzyme or substrate concentration will

Chemistry
1 answer:
lapo4ka [179]3 years ago
5 0

Answer: It will increase the rate of reaction

Explanation:

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Which celestial object is found in kuiper belt
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Water's heat of fusion is 80. cal/g , its specific heat is 1.0calg⋅∘C, and its heat of vaporization is 540 cal/g . A canister is
Pani-rosa [81]
<span>294400 cal The heating of the water will have 3 phases 1. Melting of the ice, the temperature will remain constant at 0 degrees C 2. Heating of water to boiling, the temperature will rise 3. Boiling of water, temperature will remain constant at 100 degrees C So, let's see how many cal are needed for each phase. We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion. 80 cal/g * 320 g = 25600 cal Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C 420 * 100 = 42000 cal Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away. 420 g * 540 cal/g = 226800 cal So the total number of cal used is 25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
6 0
3 years ago
1.5mol C3H8 from C3H8+5O2--&gt;3CO2+4H2O .how many grams of carbon dioxide are produced
koban [17]

Answer:

\large \boxed{\text{200 g CO}_{{2}}}

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ:                                 44.01

            C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol:    1.5

1. Calculate the moles of CO₂

The molar ratio is 3 mol  CO₂:1 mol C₃H₈

\rm  \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}

2. Calculate the mass of CO₂.

\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2}  \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}

3 0
2 years ago
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