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3 years ago

At STP a gas occupies a 64.6 kiloliter balloon. How many mol of the gas must be in the balloon? Write answer in a whole number.

Chemistry

1 answer:

Alex3 years ago

6 0

Answer:

2900 moles gas

Explanation:

PV =nRT => n = PV/RT

at STP ...

P = 1 Atm

V = 64.6KLiters = 6.46 x 10⁴ Liters

n = unknown

R = 0.08206 L·Atm/mol·K

T = 0⁰C = 273K

n = (1 Atm)(6.46 x 10⁴L)/(0.08206 L·Atm/mol·K)(273K) = 2897 moles ≈ 2900 moles

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2 KClO_2 produces 2 KCl+ 3O_2 when heated. If this reaction produces 82.8 g of KCl how many grams of O2 were produced?

Ad libitum [116K]

Answer:

37.046 grams of oxygen gas were produced.

Explanation:

$2KClO_2\rightarrow 2KCl+ 3O_2$

Moles of potassium chlorite = $\frac{82.2 g}{106.5 g/mol}=0.7718 mol$

According to reaction 2 moles of potassium chlorite gives 3 moles of oxygen gas.

Then 0.7718 moles of potassium chlorite will give:

$\frac{3}{2}\times 0.7718 mol=1.1577 mol$ of oxygen gas.

Mass of 1.1577 moles of oxygen gas:

1.1577 mol × 32 g/mol = 37.046 g

37.046 grams of oxygen gas were produced.

4 0

4 years ago

Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed

masya89 [10]

(A) $7.28\cdot 10^{-10} m$

The De Broglie wavelength of an electron is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the electron

The electron in this problem has a speed of

$v=1.00\cdot 10^6 m/s$

and its mass is

$m=9.11\cdot 10^{-31} kg$

So, its momentum is

$p=mv=(9.11\cdot 10^{-31} kg)(1.00\cdot 10^6 m/s)=9.11\cdot 10^{-25}kg m/s$

And substituting into (1), we find its De Broglie wavelength

$\lambda=\frac{6.63\cdot 10^{-34}Js}{9.11\cdot 10^{-25} kg m/s}=7.28\cdot 10^{-10} m$

(B) $1.16\cdot 10^{-34}m$

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

$p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s$

And so, the De Broglie wavelength of the ball is given by

$\lambda=\frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{5.72 kg m/s}=1.16\cdot 10^{-34}m$

(C) $9.02\cdot 10^{-9}m$

The location of the first intensity minima is given by

$y=\frac{L\lambda}{a}$

where in this case we have

$y=0.492 cm = 4.92\cdot 10^{-3} m$

L = 1.091 is the distance between the detector and the slit

$a=2.00\mu m=2.00\cdot 10^{-6}m$ is the width of the slit

Solving the formula for $\lambda$ , we find the wavelength of the electrons in the beam:

$\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m$

(D) $7.35\cdot 10^{-26}kg m/s$

The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:

$p=\frac{h}{\lambda}$

where here we have

$\lambda=9.02\cdot 10^{-9}m$ is the wavelength

Substituting into the formula, we find

$p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s$

7 0

3 years ago

A 1.897g sample of Mg(HCO3)2 was heated and decomposed. When the sample

svlad2 [7]

1.5 % is the percent yield in the reaction.

Explanation:

Given that:

original mass of the sample used in reaction = 1.897 grams

product formed after decomposition = 1.071 grams

The reaction for the decomposition:

Mg(HCO3)2 (s) ⇒ CO2 (g) + H2O (g) + MgCO2 (s)

It says that 1 mole of Mg(HCO3)2 yielded 1 mole of MgCO2 on decomposition

68.31 grams/mole or 68.31 grams of MgCO2 is formed

percent yield = $\frac{actual yield}{theoretical yield}$ x 100

putting the values in the equation:

percent yield = $\frac{1.071}{68.31}$

= 0.015 x100

PERCENT YIELD = 1.5 %

8 0

3 years ago

Which type of bond will most likely be found in HBr?

NARA [144]

<span>HBr is covalent bond.!

</span>Hydrogen and Bromine will share their electrons, to obtain the electron configuration of a noble gas<span>.
</span>
hope this helps!

6 0

3 years ago

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Question 5 why does grinding solid crystals increase the rate of dissolving for a solid solute in water? smaller crystals are mo

My name is Ann [436]

Grinding solid crystals increase the rate of dissolving for a solid solute in water because smaller crystals have more surface area. The solubility of a substance depends on the physical and chemical properties of the solute and solvent as well as the temperature, pressure and the pH of the solution. For example increase in temperature increases the rate at which a solute dissolves in a solvent.

5 0

3 years ago