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2 years ago

Seawater is an example of a _______ solution

Chemistry

1 answer:

My name is Ann [436]2 years ago

3 0

Answer:

C. Solid in liquid

Explanation:

Seawater is an example of a solid in liquid solution.

Sea water is made up of:

water
mineral salts
dissolved gases

A liquid solution is always made up of a solute being dispersed within the solvent medium.

The solvent is the liquid or fluid medium.

Such solutions are homogeneous because the solute particles are distributed evenly or uniformly in the solvent.

The solute is usually present in smaller amount compared to the solvent.

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How could measuring the melting point of a solid help decide whether it was a mixture or a compound?

mars1129 [50]

Answer: It can't.

Explanation:

In most cases, the melting point alone will not enable you to identify a compound. Millions of solid organic compounds, and their melting points, are known. Perhaps 10,000 of these will have the same melting point as your unknown compound.

Hope this helps!

8 0

3 years ago

. If sulphur (IV) oxide and methane are

zubka84 [21]

Answer:

The answer is C. 1:2

Explanation:

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

Is the following reaction balanced or unbalanced?<br> 2Fe2O3 + 3C → 4Fe + 3CO2

crimeas [40]

Answer:

unbalanced

hope this helps

7 0

3 years ago

Which is a homogeneous mixture?

Klio2033 [76]

B i think but im not 100 percent sure

7 0

3 years ago