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3 years ago

13

Name 3 variables that afect a living systom

1 answer:

emmasim [6.3K]3 years ago

5 0

1. Environment
2. Temperature
3. Food

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The energy in biomass is used to make electricity. How is the energy stored in

juin [17]

Answer:A

Explanation:

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3 years ago

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PIT_PIT [208]

The melting point of pure lead : 327 °C

<h3>Further explanation</h3>

Given

Amount of tin and melting point of solder

Required

The melting point

Solution

The composition of solder = tin and lead

So if it is 100% tin, 0% lead or 0% tin, 100% lead

From the table it is shown that when the position is 100% tin in the solder, the melting point of the solder is 232 °C, so it shows that the melting point of pure lead is obtained when% tin in solder = 0 (100% lead in solder), so that the melting point is obtained. : 327 °C

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3 years ago

Why do you think the same color M&M was used in each sample of water?

Colt1911 [192]

Answer:

The candy coating is made up of coloring and sugar. The coloring and the sugar molecules both have positive and negative charges on them. The water molecule has positive and negative charges so it can attract and dissolve the color and sugar pretty well.

Explanation:

go to https://www.acs.org/content/acs/en/education/whatischemistry/adventures-in-chemistry/experiments/dissolving-m-ms.html#:~:text=The%20candy%20coating%20is%20made,color%20and%20sugar%20pretty%20well.

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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 $NH_3$ .

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3 years ago

PPPPPPPLLLLLLLLZZZZZZ HELP I'LL GIVE BRAINLIEST

Alex17521 [72]

Food web should be the correct answer

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3 years ago

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