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pantera1 [17]
3 years ago
10

A drought hits the habitat of a semi-aquatic bird population. All ponds dry up, and fish populations decline. There are two grou

ps of birds in the population that differ in leg length and diet. Long-legged birds eat fish, while short-legged birds eat insects. The drought has little effect on insect populations.
Chemistry
1 answer:
alexandr1967 [171]3 years ago
4 0

Answer:

The population of the long-legged birds decreases.

Explanation:

The population of the short-legged birds increases whereas the population of  long-legged birds decreases due to availability of food in that environment. The long-legged birds feed on fish whose population decreases due to drought conditions so the population of long-legged birds also decreases while on the other hand, the population of short-legged birds increases or remain the same because they feed on the insects and the insects are available in large amount and less affected by the drought conditions.

You might be interested in
Determine the location of the last significant place value by placing a bar over the digit.
Vlad1618 [11]

Answer:

8,040  

0.0300  

699.5  

2.000 x 102

0.90100  

90, 100  

4.7 x 10-8  

10,800,000.0  

3.01 x 1021

0.000410

Explanation:

First remember the following rules of determining the last significant place value :

1. The digits from 1-9 are all significant and zeros between significant digits are also significant.

2.  The trailing or ending zeroes are significant only in case of a decimal number otherwise they are ignored. However starting zeroes of such a number are not significant.

Now observing above rules, lets determine the location of the last significant place value of each given example. I am determining the location by turning the last significant place to bold.

1) 8,040

8,040

Location of the last significant place value is 3 and bar is over last significant digit that is 4. Number is not decimal so ending zero is ignored. Every non zero digit is a significant.

2)  0.0300

0.0300

Location is 3 and bar is over 0. Number has a decimal point so ending zero is not ignored but starting zeroes are ignored.

3) 699.5

699.5

Location is 4 and bar is over 5.

4) 2.000 x 10²

2.000 x 10²

Location is 4 and bar is over 0. This is because the number is decimal so trailing zeroes cannot be ignored. Also if we convert this number it becomes:

200.0 so last significant digit is 0 and location of last significant digit is 4.

5) 0.90100

0.90100

Location is 5 and bar is over 0. This is because in a number with decimal point starting zeroes are ignored but trailing zeroes after decimal point are not ignored. So we count from 9 and last significant digit is 0.

6) 90, 100

90, 100

Location is 3 and bar is over 1. This is because it is not a number with decimal point. So the trailing zeroes are ignored. The count starts from 9 and last significant is 1.

7) 4.7 x 10⁻⁸

4.7 x 10⁻⁸

Location is 2 and bar is over 7. This is because the starting zeroes in a number with a decimal point are ignored. So the first digit considered is 4 and last significant digit is 7. If we expand this number:

4.7 x 10⁻⁸ =    0.000000047 = 0.000000047

Here the starting zeroes are ignored because there is a decimal point in the number.

8) 10,800,000.0

10,800,000.0

Location is 9 and bar is over 0.  Number has a decimal point so ending zero is not ignored and last significant figure is 0.

However if the number is like:

10,800,000. Then location would be 8 and bar is over 0.

9) 3.01 x 10²¹

3.01 x 10²¹

Location is 3 and bar is over 1. Lets expand this number first

3.01 x 10²¹ = 3.01 x 1000000000000000000000

                  = 3010000000000000000000

So this is the number:

3010000000000000000000

Since this is number does not have a decimal point so the trailing zeroes are ignored. Hence the count starts from 3 and the last significant figure is 1

10) 0.000410

0.000410

Location is 3 and bar is over 0. This is because the number has a decimal point so the ending zero is not ignored but the starting zeroes are ignored according to the rules given above. Hence the first significant figure is 4 and last significant figure is 0.

6 0
3 years ago
Scientists have been studying four populations in one area of the Atlantic Ocean. In this area, both tuna and dolphins eat squid
IrinaVladis [17]

Answer:

the squid population could decrease, because the tuna population is increasing and they would eat more squid, so the dolphin population would also decrease because there would be less squid to eat, and because there are less squid the herring population would increase.

3 0
3 years ago
Which observation provides evidence that some kinds of mediums can bend
Lisa [10]

Answer:

im pretty sure its A, if not so sorry!!

Explanation:

6 0
3 years ago
Read 2 more answers
The empirical formula of a compound is ch. it's molar mass is 52.07 g/mole. What is its molecular formula
Rina8888 [55]
For this item, we find first the mass of the compound with the empirical formula. That is,
                             molar mass = 12 (for C) + 1 (for H)
                                                   = 13 g/mole
Then, we divide the given mass of the molecular formula by the mass of the empirical formula. 
                                    n = 52.07 / 13 = 4
Therefore, the molecular formula is C₄H₄.
8 0
4 years ago
Soda ash (sodium carbonate) is widely used in the manufacture of glass. Prior to the environmental movement much of it was produ
egoroff_w [7]

Answer:

NaCl is the limiting reactant

18.0g of CaCO3 are in excess

113.4g Na2CO3 are produced

118.7g CaCl2 are produced

Explanation:

To solve this question we need to convert the mass of each reactant to moles in order to find the limitng and excess reactant to find the amount of products that would be produced:

<em>Moles CaCO3:</em>

125g * (1mol / 100.09g) = 1.25 moles

<em>Moles NaCl:</em>

125g * (1mol / 58.44g) = 2.14 moles

For a complete reaction of 2.14 moles of NaCl are required:

2.14 moles NaCl * (1mol CaCO3 / 2 mol NaCl) = 1.07 moles of CaCO3.

As there are 1.25 moles, <em>Calcium carbonate is the excess reactant and NaCl the limiting reactant</em>

The moles of CaCO3 in excess are:

1.25mol - 1.07mol = 0.18mol CaCO3 and the mass is:

0.18mol CaCO3 * (100.09g / mol) = 18.0g of CaCO3 are in excess

The moles of Na2CO3 and CaCl2 produced are:

2.14 moles NaCl * (1mol Na2CO3-1mol CaCl2 / 2 mol NaCl) = 1.07 moles are produced.

The masses are:

1.07 mol Na2CO3 * (105.99g / mol) = 113.4g Na2CO3

1.07 mol CaCl2 * (110.98g / mol) = 118.7g CaCl2

8 0
3 years ago
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