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2 years ago

What are three ways invasive species can be introduced into an environment?

Chemistry

1 answer:

Naily [24]2 years ago

6 0

Answer:

Explanation:

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Balance this equation: Al+ HNO3 ---- H2 + Al(NO3)3

romanna [79]

Answer:

2Al+ 6HNO3 ---- 3H2 + 2Al(NO3)3

Explanation:

Put coefficient a,b,c, and d for calculation:

a Al + b HNO3 = c H2 + d Al(NO3)3

for Al: a = d

for H: b = 2c

for N: b = 3d

for O: 3b = 9d

Suppose a=1, then d=1, b=3, c=3/2

multiply 2 to make all natural number, a=2, then b=6, c=3, d=2

6 0

3 years ago

A precipitate is a solid that sometimes forms when two liquids combine. A true B False

r-ruslan [8.4K]

<h2>Question:</h2>

A precipitate is a solid that sometimes forms when two liquids combine.

<h2>Answer:</h2>

A. True 

<h2>Explanation:</h2>

Because the Precipitate it's forms solid when two liquids combine to precipitate.

<h2>#CARRYONLEARNING </h2><h2>#STUDYWELL </h2>

3 0

3 years ago

Read 2 more answers

Look at the following reaction. How could you increase the production of 2NH3(g)? N2(g) + 3H2(g) 2NH3(g) Increase the volume of

Zolol [24]

The correct option is B. To increase the production of ammonia, you have to increase the pressure of the system. Increase in pressure will result in increased production of ammonia because this will drive the chemical reaction forward.

5 0

3 years ago

Read 2 more answers

What is responsible for the difference in climate between eastern and western California

PilotLPTM [1.2K]

I believe that the answer is D.

I hope this helps. :)

7 0

3 years ago

Calculate the theoretical yield for the bromination of both stilbenes

podryga [215]

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

For this data:

moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g

cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g

trans-stilbene

moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols

Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

4 0

2 years ago