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2 years ago

A Flame test is an example of chemistry property. What evidence can you use to back your statement up?

1 answer:

7 0

The flame test is a qualitative test used in chemistry to help determine the identity or possible identity of a metal or metalloid ion found in an ionic compound. If the compound is placed in the flame of a gas burner, there may be a characteristic color given off that is visible to the naked eye. And for the proof. The flame test provided evidence that specific atoms are present in compounds by the color of the flame. The metal atoms are what is responsible for the colors during the flame test. The color of the flame will be yellow-orange because Sodium (Na) is present in all the compounds that have a yellow-orange flame. Hope this helps! Mark brainly please!

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The message that a neuron carries

Vinvika [58]

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3 years ago

How many oxygen molecules are required for glycolysis

morpeh [17]

Answer:

Explanation:

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2 years ago

Read 2 more answers

Using stoichiometry, determine the mass of powdered drink mix needed to make 1.0 M solution of 100 mL

Rzqust [24]

Answer:

34.23 g.

M = (no. of moles of solute)/(V of the solution (L)).

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3 years ago

How many grams of KBr is required to prepare 100 mL of<br> 2.0 M KBr solution?

sesenic [268]

Answer:

23.8g

Explanation :

Convert 2.0M into mol using mol= concentration x volume

2.0M x 0.1L (convert 100mL to L since the units for M is mol/L)

= 0.2 mol

We can now find grams by using the molar mass of KBr

=119.023 g/mol (Found online) webqc.org

but can be be calculated by using the molecular weight of K and Br found on the periodic table

We can now calculate the grams by using grams=mol x molar mass

119.023g/mol x 0.2mol

= 23.8046 g

=23.8g (rounded to 1decimal place)

4 0

3 years ago

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.

Adding all three together:

$\text{A} \rightarrow 2\text{C}+\text{E}$ ( $\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 }$ )

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

4 0

1 year ago