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gogolik [260]
3 years ago
8

PLEASE HELP

Chemistry
1 answer:
WITCHER [35]3 years ago
4 0
The right answer is high number of offspring produced. asexual reproduction only requires on orgainism and creates an identical copy leaving no room for an increase of intelligence. plus it creates much more offspring and more often than sexual reproduction.
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1. What is the name for the material that stacks around the edges of impact craters on the moon?
zalisa [80]

Answer:

regolith

Explanation:

3 0
3 years ago
Write the complete electron configurations with spaces between the sublevels. If you do not format correctly your answers will a
Vinil7 [7]

Answer:

O

2−

:

Explanation:

The atomic number of O=8=number of electrons

Number of electrons in O

2−

are 8+2=10

Electronic configuration will be 1s

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2s

2

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6

3 0
4 years ago
An insulated tank having a total volume of 0.6 m3 is divided into two compartments. Initially one compartment contains 0.4 m3 of
Bess [88]

Answer:

See explaination

Explanation:

In order to have the detailed and step by step solution of the given problem, check or see the attached files.

3 0
3 years ago
Assuming that you start with 21.4 g of ammonia gas and 18.0 g of sodium metal and assuming that the reaction goes to completion,
Norma-Jean [14]

Answer:

m_{NaNH_2}=30.42gNaNH_2

m_{H_2}=0.783gH_2

Explanation:

Hello,

In this case, the reaction between sodium and ammonia is:

2Na+2NH_3\rightarrow 2NaNH_2+H_2

Thus, as we know the initial masses of both sodium and ammonia, we should first identify the limiting reactant, for which we firstly compute the available moles of sodium:

n_{Na}=18.0gNa*\frac{1molNa}{23.0gNa}=0.783molNa

And the moles of sodium consumed by 21.4 g of ammonia (2:2 mole ratio):

n_{Na}^{\ consumed}=21.4gNH_3*\frac{1molNH_3}{17gNH_3} *\frac{2molNa}{2molNH_3} =1.26molNa

In such a way, since less moles of sodium are available than consumed by ammonia, we can say, sodium is the limiting reactant. Furthermore, the mass of both sodium amide (39 g/mol) and hydrogen gas (2 g/mol) that are produced turn out:

m_{NaNH_2}=0.783molNa*\frac{2molNaNH_2}{2molNa}*\frac{39gNaNH_2}{1molNaNH_2}=30.42gNaNH_2

m_{H_2}=0.783molNa*\frac{1molH_2}{2molNa}*\frac{2gH_2}{1molH_2}=0.783gH_2

Best regards.

3 0
3 years ago
How do you calculate the rate of a reaction?
NeTakaya
Rate of reaction

The rate of a reaction is a measure of how quickly a reactant is used up, or a product is formed.

There are different ways to determine the rate of a reaction. The method chosen usually depends on the reactants and products involved, and how easy it is to measure changes in them.

The mean rate of reaction can be calculated using either of these two equations:


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Measuring mass

The change in mass of a reactant or product can be followed during a reaction. This method is useful when carbon dioxide is a product which leaves the reaction container. It is not suitable for hydrogen and other gases with a small relative formula mass, Mr.

The units for rate are usually g/s or g/min.
6 0
2 years ago
Read 2 more answers
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