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3 years ago

Which organisms are producers in a temperate coniferous forest? (Select all that apply.)

Chemistry

2 answers:

viva [34]3 years ago

7 0

Grasses. Hope this helps!! Have a great day

Tju [1.3M]3 years ago

4 0

Answer:Coniferous trees, like pine, spruce, and fir, are the main producers in the forest. Their pine needles fall to the forest floor, creating a spongy mat for other life to inhabit. Low lying shrubs and moss are other producers in this ecosystem. Producers in the coniferous forest.

Explanation:

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Sodium's atomic number is 11. What does this tell you about an atom of sodium?

Dafna1 [17]

This tells us that an atom of sodium contains 11 electrons which is balanced by 11 protons. The atomic mass of the atom is then the sum of the protons and neutrons. If you know the atomic mass of sodium, you can subtract 11 (protons) to get the number of neutrons.

5 0

3 years ago

Please help me solve this!

yulyashka [42]

Answer : The image is attached below.

Explanation :

For $O_3$ :

Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n = $\frac{m}{M}=\frac{24g}{48g/mol}=0.5mol$

Number of particles, N = $n\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.0\times 10^{23}$

For $NH_3$ :

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n = $\frac{m}{M}=\frac{170g}{17g/mol}=10mol$

Number of particles, N = $n\times 6.022\times 10^{23}=10\times 6.022\times 10^{23}=6.0\times 10^{24}$

For $F_2$ :

Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n = $\frac{m}{M}=\frac{38g}{38g/mol}=1mol$

Number of particles, N = $n\times 6.022\times 10^{23}=1\times 6.022\times 10^{23}=6.0\times 10^{23}$

For $CO_2$ :

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m = $n\times M=0.10mol\times 44g/mol=4.4g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.10\times 6.022\times 10^{23}=6.0\times 10^{22}$

For $NO_2$ :

Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m = $n\times M=0.20mol\times 46g/mol=9.2g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.20\times 6.022\times 10^{23}=1.2\times 10^{23}$

For $Ne$ :

Molar mass, M = 20 g/mol

Number of particles = $1.5\times 10^{23}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.5\times 10^{23}}{6.022\times 10^{23}}=0.25mol$

Mass, m = $n\times M=0.25mol\times 20g/mol=5g$

For $N_2O$ :

Molar mass, M = 44 g/mol

Number of particles = $1.2\times 10^{24}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.2\times 10^{24}}{6.022\times 10^{23}}=1.9mol$

Mass, m = $n\times M=1.9mol\times 44g/mol=83.6g$

For unknown substance:

Number of particles = $3.0\times 10^{23}$

Mass, m = 8.5 g

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{3.0\times 10^{23}}{6.022\times 10^{23}}=0.50mol$

Molar mass, M = $\frac{m}{n}=\frac{8.5g}{0.50mol}=17g/mol$

The substance is $NH_3$ .

3 0

3 years ago

Redox reactions can be written as two half-reactions, focusing on the gain or loss of electrons by one of the chemical substance

lutik1710 [3]

Answer:

C) H⁺

Explanation:

When we are balancing the reaction in an acid medium, hydrogen is balanced using the H⁺ species. This is most likely the intended answer of your question.

When the reaction takes place not in an acid medium, but in an alkaline one, then hydrogen is present as the OH⁻ species. However this option is not given in your question.

Thus the answer is option C).

5 0

4 years ago

DNA coils tightly during division and assembles into visible..

Alenkinab [10]

the answer is chromosomes

4 0

3 years ago

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Calculate to three significant digits the density of boron trifluoride gas at exactly 20 C and exactly 1atm . You can assume bor

Fed [463]

Answer: