How many moles of Br atoms are in .5 mol of Br2 molecules

Chemistry help please! I just need to make sure the answers correct.... Thank you!

Veronika [31]

1. At constant tempaerature and pressure, 3 tablets produce 600cm^3 of gas
Thus calculating for 1 tablet that produces 600 / 3 = 200 cm^3
So now two tablets produce 200 x 2 = 400 cm^3
2. We have the equation PV = nRT, n being the number of moles
Pressure P = 1,000 kPa
Volume V = 3 L
R = 8.31 L kPa/mol-K
Temperature T = 298 K
n = PV / RT = (1000 x 3) / (8.31 x 298) = 3000 / 2476.38 = 1.21 moles
Number of moles = 1.21 moles.

8 0

3 years ago

The density of ethanol is 0.789 g/cm^3 at 20 degrees C. What volume of ethanol has a mass of 15.5 grams?

lys-0071 [83]

<span>density = mass / volume

given the quotient , we have density and mass, volume can be easily calculated as:

volume = mass / p =15.5 g / 0.789 g/cm^3

=~ 20 cm^3 (dimension ally constant)</span>

6 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

2 years ago

Reaction rate is expressed in terms of changes in the concentration of reactants and products. Write a balanced equation for the

KengaRu [80]

Answer : The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

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.

$Rate=-\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{d[CO_2]}{dt}$

The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

5 0

3 years ago

What's the volume of one mole of an ideal gas at Standard Temperature and Pressure? Question 17 options: A) 11.2 L B) 22.4 L C)

mel-nik [20]

Answer:

=zero degrease, 1 atm)? so the volume of an ideal gas is 22.l/mol at STP this, 22l.4Lis probably the most remembered and least useful number in chemistry

4 0

2 years ago