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ss7ja [257]
3 years ago
8

Pleasee help me I really need an answer help

Chemistry
1 answer:
Juliette [100K]3 years ago
6 0

Answer:

This is hard for me also but I am trying to find out answer

Explanation:

Please mark me as brainlist.

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Explanation:

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6 0
3 years ago
Calculate the wavelength of light with a frequency of 1.89 x 1018 Hz.
Charra [1.4K]
Which sentence from "Black Hole Beginnings" contains a cause-and-effect relationship?
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3 years ago
Enzymes work by _____. enzymes work by _____. decreasing the potential energy difference between reactant and product adding ene
Anton [14]
The appropriate answer is they decrease the potential energy difference between reactant and product. They do this by bringing products and reactants together at the active site on the enzyme molecule. Enzymes are biological catalysts. They increase the rate of a chemical reaction without being used up in the reaction itself. So one molecule of an enzyme is used many times to catalyze the reaction of other molecules. 
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3 years ago
What is a covalent compound? physical science?
umka21 [38]
Covalent compounds are formed when non metals bond together.
4 0
3 years ago
Suppose 6.63g of zinc bromide is dissolved in 100.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final mo
Alenkasestr [34]

Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]

We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:

<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>

4 0
3 years ago
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