Lucky charms A. Element B. Compound C.Homogeneous Solution D.Heterogeneous

4. Given the balanced equation: 2Na + S → Na₂S

PtichkaEL [24]

Answer:

Option D. 30 g

Explanation:

The balanced equation for the reaction is given below:

2Na + S —> Na₂S

Next, we shall determine the masses of Na and S that reacted from the balanced equation. This is can be obtained as:

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g

Molar mass of S = 32 g/mol

Mass of S from the balanced equation = 1 × 32 = 32 g

SUMMARY:

From the balanced equation above,

46 g of Na reacted with 32 g of S.

Finally, we shall determine the mass sulphur, S needed to react with 43 g of sodium, Na. This can be obtained as follow:

From the balanced equation above,

46 g of Na reacted with 32 g of S.

Therefore, 43 g of Na will react with = (43 × 32)/46 = 30 g of S.

Thus, 30 g of S is needed for the reaction.

8 0

3 years ago

1. What are the possible genotypes of offspring when crossing BB, Bb parents?

marissa [1.9K]

Answer:

1, C: BB, Bb, Bb, BB

2. C: Hybrid

Explanation:

1. If u do a punnet square for BB and Bb you will get: BB, Bb, Bb, Bb

B| B|

B| BB. BB

b| Bb Bb

2. You do a punnet square for BB and bb and you'll get: Bb, Bb, Bb, Bb, which means all the children are hybrids of Dominant alleles ans recessive alleles

B  B

b| Bb Bb

4 0

3 years ago

Read 2 more answers

23. Using the periodic table, what is the oxidation number of:

Aloiza [94]

Answer:

a. +2

b. +3

c. -1

Explanation:

The typical oxidation states can be determined from the periodic table based on the number of valence electrons an atom has.

a. Calcium belongs to group 2A, meaning it has 2 valence electrons and, therefore, would have an oxidation state of +2 in compounds.

b. Aluminum is in group 3A, meaning it has 3 valence electrons and would have an oxidation state of +3 in compounds when the 3 electrons are lost.

c. Fluorine would become fluorine if it gained 1 additional electron to achieve an octet, so its oxidation state would be -1.

8 0

4 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

For the reaction N2O4(g) ⇋ 2NO2(g), Kc = 0.25 at 98°C. At a point during the reaction, the concentration of N2O4 = 0.50 M and th

Scilla [17]

Answer:

Q = 0.50

No

Left

Explanation:

At a generic reversible equation

aA + bB ⇄ cC + dD

The reaction coefficient (Q) is the ratio of the substances concentrations:

$Q = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}$

Solids and liquid water are not considered in this calculus.

When the reaction achieves equilibrium (concentrations are constant), the Q value is named as Kc, which is the equilibrium constant of the reaction. If Q > Kc, it indicates that the concentration of the products is higher, so, the reaction must progress to the left and form more reactants; if Q < Kc, than the concentrations of the reactants, are higher, so, the reaction progress to the right.

In this case:

Q = $\frac{[NO_2]^2}{[N_2O_4]}$

$Q = \frac{0.50^2}{0.50}$

Q = 0.50

So, Q > Kc, the reaction is not at equilibrium and it progresses to the left.

6 0

3 years ago

Lucky charms A. Element B. Compound C.Homogeneous Solution D.Heterogeneous​

Lucky charms A. Element B. Compound C.Homogeneous Solution D.Heterogeneous