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3 years ago

Where do reactions in a solid generally take place?

Chemistry

1 answer:

Sholpan [36]3 years ago

3 0

Answer:

It's D. On the surface of the solid.

Explanation:

If the reactant is a solid, the surface area of the solid will impact how fast the reaction goes. This is because the two types of molecule can only bump into each other at the liquid solid interface, i.e. on the surface of the solid. So the larger the surface area of the solid, the faster the reaction will be.

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An unknown compound has a percent composition of 52.10% potassium, 15.8% carbon, and 32.1% oxygen. The molar mass of the compoun

lianna [129]

To determine the empirical formula and the molecular formula of the compound, we assume a basis of the compound of 100 g. We do as follows:

Mass Moles
K 52.10 52.10/39.10 = 1.33 1.33/1.32 ≈ 1
C 15.8 15.8/12 = 1.32 1.32/1.32 ≈ 1
O 32.1 32.1 / 16 = 2.01 2.01/1.32 ≈ 1.5

The empirical formula would most likely be KCO.
The molecular formula would be K2C2O3.

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seropon [69]

The labeled diagram is given in the image attached.

As it can be seen from the image that freezing is when energy is removed from the system at 0 ⁰ while melting is when energy is added at 0⁰.

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3 years ago

Which subatomic particle of an atom never changes?

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3 years ago

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

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3 years ago