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3 years ago

13

Convert 2,247,896 milligrams (mg) to pounds (lb) rounding to 3 sig figs

1 answer:

Dovator [93]3 years ago

6 0

4.96 would be you’re answer

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What is required to cause acceleration or cause an object to move

OverLord2011 [107]

A "FORCE" is required to cause acceleration or cause an object to move.

7 0

3 years ago

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Which of the following is a direct result of hydrogen bonding?

prisoha [69]

Cohesion and adhesion

5 0

3 years ago

In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n

artcher [175]

Stoichiomety:

1 moles of C + 1 mol of O2 = 1 mol of CO2

multiply each # of moles times the atomic molar mass of the compund to find the relation is weights

Atomic or molar weights:

C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol

Stoichiometry:

12 g of C react with 32 g of O2 to produce 44 g of CO2

Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen

And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.

You cannot obtain 72 g of CO2 from 18 g of C.

May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.

3 0

4 years ago

Read 2 more answers

Given the equation:

Citrus2011 [14]

Answer: a) $Fe$ is the limiting reagent

b) 3.59 g

c) 11.6g

Explanation:

$4Al_2O_3+9Fe\rightarrow 3Fe_3O_4+8Al$

To calculate the moles :

$\text{Moles of} Al_2O_3=\frac{27.5g}{102g/mol}=0.27moles$

$\text{Moles of} Fe=\frac{8.4g}{56g/mol}=0.15moles$

According to stoichiometry :

a) 9 moles of $Fe$ require= 4 moles of $Al_2O_3$

Thus 0.15 moles of $Fe$ will require= $\frac{4}{9}\times 0.15=0.067moles$ of $Al$

Thus $Fe$ is the limiting reagent as it limits the formation of product and $Al$ is the excess reagent.

b) As 9 moles of $Fe$ give = 8 moles of $Al$

Thus 0.15 moles of $Fe$ give = $\frac{8}{9}\times 0.15=0.133moles$ of $Al$

Mass of $Al=moles\times {\text {Molar mass}}=0.133moles\times 27g/mol=3.59g$

c) As 9 moles of $Fe$ give = 3 moles of $Fe_3O_4$

Thus 0.15 moles of $Fe$ give = $\frac{3}{9}\times 0.15=0.05moles$ of $Fe_3O_4$

Mass of $Fe_3O_4=moles\times {\text {Molar mass}}=0..05moles\times 231.5g/mol=11.6g$

8 0

3 years ago

If a temperature increase from 12.0 ∘c to 22.0 ∘c doubles the rate constant for a reaction, what is the value of the activation

Alex73 [517]

Answer: 48,501 J/mol

Explanation:

1) Action barrier = activation energy = Ea

2) Data:

i) T₁ = 12°C = 12 + 273.15 K = 285.15K

ii) T₂ = 22°C = 22 + 273.15 K = 295.15 K

iii) rate constant = k: k₂ / k₁ = 2

iv) Ea = ?

3) Formula:

Arrhenius' law gives the relationship between the constant of reaction and the temperature:

$k=Ae^{-\frac{Ea}{RT}}$

4) Solution

By arranging the formula, you get:

㏑[k₂/k₁] =Ea/R [1/T₁ - 1/T₂]

Replace k₂ = 2k₁; T₁ = 285.15; and T₂ = 295.15

ln[2] = Ea/8.314 J/K mol × [1/285.15 - 1/295.15]K

Ea = ln [2] × 8.314 J/K mol / [1.18818×10⁻⁴K] = 48,501 J/mol

6 0

4 years ago