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4 years ago

15

If you are given an unknown liquid that is 1.0 L and has the mass of 500 grams which of the substance would it be. Distilled Wat

er Density= 1.0g/cm^3, Propane density 0.494 g/cm^3, Salt Water density 1.025 g/cm^3 or Liquid Gold 17.31 g/cm^3?

1 answer:

kodGreya [7K]4 years ago

8 0

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Ammonia (NH3) can be produced by the reaction of hydrogen gas with nitrogen gas:

VARVARA [1.3K]

Answer:

Percent yield = 61.5%

Explanation:

Given data:

Moles of hydrogen = 2.00 mol

Actual yield of ammonia= 0.80 mol

Percent yield = ?

Solution:

Chemical equation:

3H₂ + N₂ → 2NH₃

Now we will compare the moles of ammonia with hydrogen from balance chemical equation:

H₂ : NH₃

3 : 2

2 : 2/3×2 = 1.3 mol

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 0.80 mol /1.3 mol × 100

Percent yield = 0.615 × 100

Percent yield = 61.5%

6 0

3 years ago

if a gas is pumped from a smaller container to a container that is twice size, what happens to the pressure of the gas?

Vaselesa [24]

Answer:

The pressure of the gas would decrease

Explanation:

8 0

3 years ago

Consider the reaction NOBr(g) => NO(g) + 1/2 Br2(g). A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M-1

timama [110]

Answer:

Thus, the order of the reaction is 2.

The rate constant of the graph which is :- 2.00 M⁻¹s⁻¹

Explanation:

The kinetics of a reaction can be known graphically by plotting the concentration vs time experimental data on a sheet of graph.

The concentration vs time graph of zero order reactions is linear with negative slope.

The concentration vs time graph for a first order reactions is a exponential curve. For first order kinetics the graph between the natural logarithm of the concentration vs time comes out to be a straight graph with negative slope.

The concentration vs time graph for a second order reaction is a hyberbolic curve. Also, for second order kinetics the graph between the reciprocal of the concentration vs time comes out to be a straight graph with positive slope.

Considering the question,

A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M⁻¹s⁻¹.

<u>Thus, the order of the reaction is 2.</u>

<u>Also, slope is the rate constant of the graph which is :- 2.00 M⁻¹s⁻¹</u>

5 0

4 years ago

Be sure to answer all parts. Write the balanced equations corresponding to the following rate expressions: a) rate = − 1 3 Δ[CH4

Alinara [238K]

Answer : The balanced equations will be:

(a) $3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $2N_2O_5\rightarrow 2N_2+5O_2$

(c) $2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

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.

Now we have to determine the balanced equations corresponding to the following rate expressions.

(a) $Rate=-\frac{1}{3}\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[H_2O]}{dt}=-\frac{d[CO_2]}{dt}=+\frac{1}{4}\frac{d[CH_3OH]}{dt}$

The balanced equations will be:

$3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $Rate=-\frac{1}{2}\frac{d[N_2O_5]}{dt}=+\frac{1}{2}\frac{d[N_2]}{dt}=+\frac{1}{5}\frac{d[O_2]}{dt}$

The balanced equations will be:

$2N_2O_5\rightarrow 2N_2+5O_2$

(c) $Rate=-\frac{1}{2}\frac{d[H_2]}{dt}=-\frac{1}{2}\frac{d[CO_2]}{dt}=-\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2CO_3]}{dt}$

The balanced equations will be:

$2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

4 0

3 years ago

Which of the following factors would reduce the yield of a reaction?

Olenka [21]

Answer:

B. The reactants form additional unexpected products.

Explanation:

7 0

3 years ago