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koban [17]
3 years ago
9

What type of rock is the Wave made of?

Chemistry
1 answer:
Arada [10]3 years ago
5 0

What type of rock is the wave made of?

Answer: The Wave is a natural formation of ~sandstone rocks~

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When asked to classify sodium acetate (NaCH3 COO) as either an ionic or covalent compound, a student responded with, “Sodium ace
maxonik [38]
 Sodium acetate<span> contains </span>both ionic and covalent<span> bonds. </span> <span>In </span>sodium acetate<span>, there is </span>ionic<span> bond between sodium and acetate </span>ions<span> and there is </span>covalent<span> bond in the atoms that are bondeed to the carbon atoms. Hope this answers the question.</span>
4 0
3 years ago
Help me understand molarity please
Katyanochek1 [597]
Molarity/ molar is a unit of concentration, symbolized by "M". It is the ratio of the number of moles of solute and the volume of solution.

Molarity = Moles of solute ÷ Liters of solution = moles/liter = M
molarity is define as the number of mole of a solute in 1 liter of solution
3 0
3 years ago
Which list contains all pure substance
Viktor [21]
The periodic table contains the pure substances
4 0
3 years ago
Write the equilibrium-constant expression for the reactionA(s)+3B(l)↽−−⇀2C(aq)+D(aq)A(s)+3B(l)↽−−⇀2C(aq)+D(aq)in terms of [A], [
Ilia_Sergeevich [38]

Answer:

K_{c} = [\text{C}]^{2}[\text{[D]}

Explanation:

\rm A(s)+3B(l) \, \rightleftharpoons \, 2C(aq)+D(aq)

The general formula for an equilibrium constant expression is

K_{c} = \dfrac{[\text{Products}]}{[\text{Reactants}]}

Solids and liquids are not included in the equilibrium constant expression.

Thus, for this reaction,  

K_{c} = [\textbf{C}]^{\mathbf{2}}\textbf{[D]}

3 0
3 years ago
At what substrate concentration would an enzyme with a K_cat of 30.0 s⁻¹ and a Km of 0.0050 M operate at one-quarter of its maxi
Umnica [9.8K]

Answer:

The correct answer is option B.

Explanation:

Michaelis–Menten 's equation:

v=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}

V_{max}=k_{cat}[E_o]

v = rate of formation of products

[S] = Concatenation of substrate = ?

[K_m] = Michaelis constant

V_{max}= Maximum rate achieved

k_{cat} = Catalytic rate of the system

E_o = initial concentration of enzyme

We have :

v=\frac{V_{max}}{4}

[S] =?

K_m=0.0050 M

v=V_{max}\times \frac{[S]}{(K_m+[S])}

\frac{V_{max}}{4}=V_{max}\times \frac{[S]}{(0.0050 M+[S])}

[S]=\frac{0.005 M}{3}=1.7\times 10^{-3} M

So, the correct answer is option B.

8 0
3 years ago
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