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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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HELP!!!! please omg

OverLord2011 [107]

Answer:

I believe it's A. to reduce air bubbles. Tbh, it's been a while

7 0

3 years ago

A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0

OleMash [197]

Answer: The mass of original oxalic acid sample is 6.75 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2C_2O_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of oxalic acid = ? g

Molar mass of oxalic acid = 90 g/mol

Molarity of solution = 0.075 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g$

Hence, the mass of original oxalic acid sample is 6.75 grams

7 0

3 years ago

True or False: If this was a cooling

denpristay [2]

Answer:

True

Explanation:

5 0

3 years ago

A galvanic cell at a temperature of is powered by the following redox reaction: 2Cr3 + 3Ca Suppose the cell is prepared with in

stira [4]

Answer:

2.13 V

Explanation:

The balanced equation of the reaction his;

2Cr^3+(aq) + 3Ca(s) -----> 2Cr(s) + 3Ca^2+(aq)

Since this is a galvanic cell then E°cell must be positive. It implies that calcium will be the anode and chromium will be the cathode since calcium is ahead of chromium in the electrochemical series.

E°anode= -2.87 V

E°cathode= -0.74 V

E°cell= E°cathode -E°anode

E°cell= -0.74 -(-2.87)

E°cell = 2.13 V

8 0

3 years ago

A 160g radioactive sample is left undisturbed for 12.5 hours. At the end of that period, only 5.0g remain. What is the half-life

11Alexandr11 [23.1K]

it would be 26 because once you divide 160g by 5.0g and add 12.5 you would get 26

hope this helps

3 0

3 years ago

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