Stars form inside giant clouds of gas and dust. What are these giant clouds of gas and dust called?

The limiting reactant, O2, can form up to 2.7 mol Al2O3. What mass of Al2O3 forms?

Margaret [11]

Answer:

280 g Al₂O₃

Explanation:

To find the mass, you need to multiply the given value by the molar mass. This will cause the conversion because the molar mass exists as a ratio; technically, the ratio states that there are 101.96 grams per every 1 mole Al₂O₃. It is important to arrange the ratio in a way that allows for the cancellation of units. In this case, the desired unit (grams) should be in the numerator. The final answer should have 2 sig figs to reflect the given value (2.7 mol).

Molar Mass (Al₂O₃): 101.96 g/mol

2.7 moles Al₂O₃ 101.96 g
------------------------ x ------------------- = 275 g Al₂O₃ = 280 g Al₂O₃
1 mole

5 0

2 years ago

Answer my my science homework please

ludmilkaskok [199]

1. D

2. 83.5 million years ago - 66 million years ago

3. It had armor and could rotate to keep its caudal bludgeon facing the enemy.

hope this helped :)
alisa202

5 0

3 years ago

The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism

MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

Where K' = K1 * K2

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just mechanism A and B are consistent with observed rate law

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just mechanism A is consistent with the observation of J. H. Sullivan

5 0

3 years ago

PLS HELP QUICK ALOTTT OF POINTS

timofeeve [1]

Answer:

$\boxed {\boxed {\sf 0.80 \ mol\ F}}$

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

$\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}$

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

$4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}$

Flip the ratio so the units of atoms of fluorine cancel each other out.

$4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}$

$4.8 \times 10^{23} *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }$

Condense into 1 fraction.

$\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F$

Divide.

$0.7970773829 \ mol \ F$

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

$0.80 \ mol \ F$

4.8 × 10²³ fluorine atoms are equal to 0.80 moles of fluorine.

6 0

3 years ago

How do chemists solve problems?

Westkost [7]

Answer:Chemistry problems can be solved using a variety of techniques.

Explanation: Many chemistry teachers and most introductory chemistry texts illustrate problem solutions using the factor-label method. ... The use of analogies and schematic diagrams results in higher achievement on problems involving moles, stoichiometry, and molarity. Hope this helped!

4 0

3 years ago