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3 years ago

It is possible to increase a reaction rate by adding very small amounts of a catalyst, and this is because...

Chemistry

1 answer:

Harlamova29_29 [7]3 years ago

3 0

A. The catalyst increases the rate of reaction without being consumed

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A certain automobile contains 4 tires, 2 headlights, 1 steering wheel,

Step2247 [10]

Answer:

143 parts I think probably not

5 0

2 years ago

7. List all of the variables that you controlled or kept the same. Salt water

satela [25.4K]

Answer:

Temperature is a common type of controlled variable. If a temperature is held constant during an experiment, it is controlled. Other examples of controlled variables could be an amount of light, using the same type of glassware, constant humidity, or duration of an experiment.

Explanation:

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2 years ago

What mass of nitrogen is needed to fill an 855 L tank at STP?

scoundrel [369]

Answer:

It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C or 273.15 °K are used and are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

So, in this case:

P= 1 atm
V= 855 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 273.15 K

Replacing:

1 atm* 855 L= n* 0.082 $\frac{atm*L}{mol*K}$ * 273.15 K

Solving:

$n=\frac{1 atm* 855 L}{0.082\frac{atm*L}{mol*K} *273.15 K }$

n= 38.17 moles

Being the molar mass of nitrogen N2 equal to 28 g / mol, you can apply the following rule of three: if there are 28 grams in 1 mole, how much mass is there in 38.17 moles?

$mass=\frac{38.17 moles*28 grams}{1 mole}$

mass= 1,068.76 grams

It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.

3 0

3 years ago

An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal was formed and 5.20 L of oxygen was released at

Yakvenalex [24]

Answer:

68.3g

Explanation:

1. Word equation:

mercury(II) oxide → mercury + oxygen

2. Balanced molecular equation:

2HgO → 2Hg + O₂(g)

3. Mole ratio

Write the ratio of the coefficients of the substances that are object of the problem:

$2molHgO/1molO_2$

4. Calculate the number of moles of O₂(g)

Use the equation for ideal gases:

$pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol$

5. Calculate the number of moles of HgO

$\dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO$

6. Convert to mass

mass = # moles × molar mass

molar mass of HgO: 216.591g/mol

mass = 0.315mol × 216.591g/mol = 68.3g

7 0

2 years ago

What types of elements form covalent bonds?

spayn [35]

Covalent bonds form when electrons are shared between two nonmetals.

3 0

2 years ago