Organize the following metals from least metallic to most metallic :barium,zinc,iron, and strontium

KOH+HBr-KBr+H2O which is the base in this reaction

krek1111 [17]

Answer:

KOH is the base in this reaction

Explanation:

HBr is the acid in this reaction.

Hope this help you. Have a nice day !

8 0

3 years ago

E85 fuel is a mixture of 85% ethanol and 15% gasoline by volume. What mass in kilograms of E85 (d = 0.758 g/mL) can be contained

alexandr402 [8]

The answer is: mass is 40.17 kilograms.

d = 0.758 g/mL; density of fuel.

V = 14.0 gal; volume.

A gallon is a unit of volume in both the US customary and imperial systems of measurement. The US gallon is defined as 231 cubic inches (3.785 liters).

1 gal = 3785.41 mL.

V = 14 gal · 3785.41 mL:

V = 52995.74 mL.

m = 52995.74 mL · 0.758 g/mL.

m = 40170.77 g; mass of fuel.

m = 40170.77 g ÷ 1000 g/kg.

m = 40.17 kg.

7 0

3 years ago

Read 2 more answers

The point in a titration at which the indicator changes is called the

irakobra [83]

Answer:

End point

Explanation:

The point at which the indicator changes color is called the endpoint. So the addition of an indicator to the analyte solution helps us to visually spot the equivalence point in an acid-base titration

#correct me if I'm wrong

keep safe and study hard

brainliest please thank you

8 0

2 years ago

One of the steps in processing some metals is smelting. what does smelting do for a metal?

vichka [17]

Hey there,
Smelting's main use is to produce a base metal from its ore

Hope this helps :))

~Top

8 0

3 years ago

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

$[S] = K_M$

So this means that equation one would now become

$[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

$[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

$[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

The rate of this reaction is mathematically defined as

$v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$ is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

$v = \frac{V_{max}[S]}{[S]} = V_{max}$

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme

7 0

3 years ago