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vodomira [7]
3 years ago
5

With term describes the group of plants that most belong to?

Chemistry
1 answer:
Sergio039 [100]3 years ago
3 0

Answer:

It is called the Phylum Bryophyta.

Explanation:

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How is a planet different from a moon? (2 points)
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A

Explanation:

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Describe how the metal probably increases the reaction rate, identify whether this is an example of homogeneous or heterogeneous
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Answer:

The metal probably increases reaction rate by either holding reactant molecules in the correct orientation to react or by weakening or breaking bonds in reactant molecules to make them more reactive.

This is an example of heterogeneous catalysis.

It is heterogeneous catalysis because the catalyst is a solid and the reactants are gases. In heterogeneous catalysis, the catalyst is in a different phase than the reactants

Explanation:

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Red giant stars defination​
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A red giant is a luminous giant star of low or intermediate mass in a late phase of stellar evolution. The outer atmosphere is inflated and tenuous, making the radius large and the surface temperature around 5,000 K or lower.

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How does waer evapourate? one question brainlyist
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water evaportaes from heat it then turns into a gas then can go into a solid one day and repeat the cycle

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4 0
3 years ago
A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur
dsp73

Answer:

12.8 g of O_{2} must be withdrawn from tank

Explanation:

Let's assume O_{2} gas inside tank behaves ideally.

According to ideal gas equation- PV=nRT

where P is pressure of O_{2}, V is volume of O_{2}, n is number of moles of O_{2}, R is gas constant and T is temperature in kelvin scale.

We can also write, \frac{V}{RT}=\frac{n}{P}

Here V, T and R are constants.

So, \frac{n}{P} ratio will also be constant before and after removal of O_{2} from tank

Hence, \frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}

Here, \frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm} and P_{after}=17atm

So, n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol

So, moles of O_{2} must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of O_{2} = 32 g/mol

So, mass of O_{2} must be withdrawn = (32\times 0.40)g=12.8g

7 0
3 years ago
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