All categories

3 years ago

Smart people help me pass my class

Chemistry

1 answer:

slega [8]3 years ago

4 0

Answer:

im not smart but here

Explanation:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.

You might be interested in

139%

GaryK [48]

<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

$2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}$

$\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}$

$\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}$

$\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}$

Based on no. of iron reacted,

$\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)$

n = m/M

$\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}$

% composition of $FeCl_3$

= $(9.75 / 10.39)^{*} 100$

= 94%

6 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago

How many grams of copper (II) oxide will react with 10 liters of hydrogen gas?

Svetradugi [14.3K]

Answer:

maybe 200g

I don't know thats just a guess

HOPE ITS TURE

8 0

2 years ago

[Ca2+] within the endoplasmic reticulum (ER) is 10-3M. Cytosolic [Ca2+] is 10-7M. Lysosomes in mammalian cells have an internal

Pani-rosa [81]

Answer:

B. Ca2+ import into the ER because it has the steeper concentration gradient

Explanation:

ΔGt = RT㏑(C₂/C₁)

where ΔGt is the free energy change for transport; R = 8.315 J/mol; T = 298 K; C₂/C₁ is ratio of concentrations inside and outside each organelle.

For Ca²⁺ import

ΔGt = 8.315 J/mol * 298 K * ㏑(10⁻³/10⁻⁷)

ΔGt= 3.42 kJ/mol

For H⁺ import

ΔGt = 8.315 J/mol * 298 K * ㏑ (10⁻⁴/10⁻⁷)

ΔGt = 2.73 kJ/mol

From the above values, ΔGt is greater for Ca²⁺ import because it has a steeper concentration gradient

8 0

3 years ago

Hellllppppppp meeee fassttttt please

Volgvan

a tick attaches itself to the skin of a dog

5 0

3 years ago