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3 years ago

A student runs two experiments with a constant-volume "bomb" calorimeter containing 1500.g of water.

Chemistry

1 answer:

muminat3 years ago

3 0

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

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Calculate the number of grams of CO that can react with 0.400 kg of Fe2O3.

aleksandr82 [10.1K]

Answer:

mass of CO = 210.42 g

mass in three significant figures = 210. g

Explanation:

Given data:

mass of Fe2O3 = 0.400 Kg

mass of CO= ?

Solution:

chemical equation:

Fe2O3 + 3CO → 2Fe + 3CO2

Now we will calculate the molar mass of Fe2O3 and CO.

Molar mass of Fe2O3 = (55.845 × 2) + (16 × 3) = 159.69 g/mol

Molar mass of CO = 12+ 16 = 28 g/mol

now we will convert the kg of Fe2O3 in g.

mass of Fe2O3 = 0.400 kg × 1000 = 400 g

number of moles of Fe2O3 = 400 g/ 159.69 g/mol = 2.505 mol

mass of CO = moles of Fe2O3 × 3( molar mass of CO)

mass of CO = 2.505 mol × 84 g/mol

mass of CO = 210.42 g

mass in three significant figures = 210. g

5 0

4 years ago

Read 2 more answers

The chief advantage of the metric system over other systems of measurement is that it ____.

OlgaM077 [116]

The chief advantage of the metric system over other systems of measurement is that it B. is in multiples of 10.

This can be seen in the picture below that shows the prefixes of the metric system.

7 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p

Tanzania [10]

Answer : The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

$\text{Rate of formation of }H=+\frac{d[H]}{dt}$

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

Given:

$-\frac{d[D]}{dt}=0.18mol/L.s$

As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0

3 years ago

*Riddle!* *Riddle!!* *Riddle!!!*

otez555 [7]

Answer:

https://www.getriddles.com/science-riddles/

Explanation:

go to this link, you will find some answers

hope it helps!

5 0

3 years ago

Law of Conservation of Mass in terms of atoms

VARVARA [1.3K]

No atoms are lost or made during the chemical reaction so the total mass of the products is equal to the total mass of the reactants. In an atom, protons and neutrons contribute to the mass and since the number of them doesn’t change, the mass doesn’t either.

4 0

3 years ago