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Aneli [31]
2 years ago
8

Plsss help i’ll give brainlist

Chemistry
1 answer:
yarga [219]2 years ago
7 0

Answer:

As the temperature of the water increases, the time needed for the dye to spread decreases. This is because the kinetic energy between the liquid particles increases, therefore helping the dye to dissolve and spread throughout the water.

Explanation:

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Can you pls tell me the word equations for all these equations​
Leokris [45]

Answer:

Below

Explanation:

Balanced form;

2C_6H_6 +15O_2 -> 12CO_2+6H_2O\\\\ Ca_3(PO_4)_2 +8C -> Ca_3P_2 +8CO\\\\2HNO_2+O_2 -> 2HNO_3\\\\Ca(OH)_2 + CO_2 -> CaC0_3+H_2O\\\\2K +Br_2 ->2KBr\\\\2NaOH+FeSO_4 -. Na_2SO_4 +Fe(OH)_2

1.Benzene + Dioxygen = Carbon Dioxide + Water

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3.Nitrous acid react with oxygen to produce nitric acid.

4.This means that the carbon dioxide and limewater react to produce calcium carbonate and water.

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6. An aqueous solution of ferrous sulphate reacts with aqueous solution of sodium hydroxide to form a precipitate of ferrous hydroxide and sodium sulphate remains in the solution.

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3 years ago
The total volume of seawater is 1.5 x 10²¹ L. Seawater contains approximately 3.5% sodium chloride by mass. At that high of a co
garri49 [273]

Answer:

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater (m_{sw}), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

m_{sw} = \rho_{sw}\cdot V_{sw}

Where:

\rho_{sw} - Density of seawater, measured in grams per liters.

V_{sw} - Volume of seawater, measured in liters.

If V_{sw} = 1.5\times 10^{21}\,L and \rho_{sw} = 1030\,\frac{g}{L}, then:

m_{sw}=\left(1030\,\frac{g}{L} \right)\cdot (1.5\times 10^{21}\,L)

m_{sw} = 1.545\times 10^{24}\,g

The total mass of sodium chloride is determined by the following ratio:

r = \frac{m_{NaCl}}{m_{sw}}

m_{NaCl} = r\cdot m_{sw}

Given that m_{sw} = 1.545\times 10^{24}\,g and r = 0.035, the total mass of sodium chloride in all the seawater in the world is:

m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)

m_{NaCl} = 5.408\times 10^{22}\,g

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

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