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3 years ago

PLEASE HELP QUICK ILL RRALLY APPRECIATE IT

Chemistry

1 answer:

geniusboy [140]3 years ago

7 0

Sugar-Compound

Salt Water- Liquid Solution

Air-Gaseous Solution

Sterling Silver- Solid Solution

Gold- Element

Hope this helps:)

If you could, would you please mark brainliest?

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What is the volume of 6.50g of gold if the density of gold is 19.3g/ml?

Jet001 [13]

Answer:

0.33 ml

Explanation:

Formula for density: p = m/V (where p is density, m is mass and V is volume)

So, by applying the above formula:

$19.3 = \frac{6.5}{v}$

$v = 0.33 \: ml$

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What are small round structures in the cell that breaks down large food particles into smaller ones

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The small particles are known as protiens

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Read 2 more answers

Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of SO2 by O2 to give SO3

SIZIF [17.4K]

Answer:

See explanation

Explanation:

We know that the process of the oxidation of SO2 to SO3 is catalysed by NO2 gas. It occurs in two stages and i will show the balanced reaction equation of the both stages below;

Step 1

2NO2(g) + 2SO2(g) ------> 2NO(g) + 2SO3(g)

Step 2

2NO(g) + O2(g) -------> 2NO2(g)

So, the overall reaction equation is;

2SO2(g) + O2(g) ------> 2SO3(g)

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3 years ago

2. (2 pts) How would you prepare 1.5 liters of 2 M KCI (MW=74.55 g/mol)

ra1l [238]

Answer:

Dissolve 226 g of KCl in enough water to make 1.5 L of solution

Explanation:

1. Calculate the moles of KCl needed

$n = \text{1.5 L} \times \dfrac{\text{2 mol}}{\text{1 L}}= \text{3.0 mol}$

2. Calculate the mass of KCl

$m = \text{3.0 mol} \times \dfrac{\text{74.55 g}}{\text{1 mol}}= \text{224 g}$

3. Prepare the solution

Measure out 224 g of KCl.
Dissolve the KCl in a few hundred millilitres of distilled water.
Add enough water to make 1.5 L of solution. Mix thoroughly to get a uniform solution.

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What is the number assigned to an element on the Periodic Table?

spayn [35]

Answer:

Element Atomic Number One number you will find on all periodic tables is the atomic number for each element.

Explanation:

hope it helps :))

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3 years ago