What's a wind up toy useful energy output

poizon [28]

<h2>Hello!</h2>

The answer is:

The empirical formula is the option B. $NH_{3}$

The empirical formula of a compound is the simplest formula that can be written. On the opposite, the molecular formula involves a variant of the same compound, but it can be also simplified to an empirical formula.

$MolecularFormula=n(EmpiricalFormula)$

We are looking for a formula that cannot be simplified by dividing the number of molecules/atoms that conforms the compound.

Let's discard option by option in order to find which formula is an empirical formula (cannot be simplified)

A. $N_{2}O_{4}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$N_{2}O_{4}=2(NO_{2})$

B. $NH_{3}$

It's an empirical formula since it cannot be obtained by the multiplication of a whole number and the simplest formula. It's the simplest formula that we can find of the compound.

C. $C_{3}H_{6}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$C_{3}H_{6}=3(CH_{2})$

D. $P_{4}O_{10}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$P_{4}O_{10}=2(P_{2}O_{5})$

Hence, the empirical formula is the option B. $NH_{3}$

Have a nice day!

6 0

3 years ago

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Cr2o2−7(aq)+i−(aq)→cr3+(aq)+io−3(aq) (acidic solution) express your answer as a chemical equation. identify all of the phases in

Rashid [163]

Answer: Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either. Start with your 2 half reactions: I- --> IO3- Cr2O72- --> 2 Cr3+ Balance O by adding H2O: I- + 3 H2O --> IO3- Cr2O72- --> 2 Cr3+ + 7H2O Balance H by adding H+: I- + 3 H2O --> IO3- + 6 H+ Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O Balance charge by adding e-: I- + 3 H2O --> IO3- + 6 H+ + 6 e- Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give: Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O

4 0

2 years ago

What is meant by the term sensitivity in Analytical Chemistry? (a) The term refers to the limit at which the absorption instrume

ivolga24 [154]

The term sensitivity in Analytical Chemistry is "the slope of the calibration curve or a function of analyte concentration or amount".

Answer: Option B

Explanation:

In a sample, the little amounts of substances can be accurately evaluated by a method is termed as "Analytical sensitivity". This detect a target analyte like an antibody or antigen, process is considered as potential of a test to and generally demonstrated as the analyte's minimum detectable concentration.

The acceptable diagnostic sensitivity is not guaranteed by high analytical sensitivity. The percentage of individuals who have a given disarray who are identified by the method as positive for the disarray is known as "Diagnostic sensitivity".

6 0

3 years ago

Question is in the image

zloy xaker [14]

Answer:(3)

Explanation: 2Al+3H2SO4----->Al(2SO4)3+3H2

6 0

3 years ago

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Identify the reversible chemical reaction below. A. reactants=>products B. Products <=> reactants C. products => rea

Korvikt [17]

Answer:

Its B

Explanation:

Products <=> reactants; products can go back to reactant and reactants can form products, depending on the equilibrium position.

The equilibrium position, if it is on the left, and much of the product is formed. If the equilibrium position is not the right, then its the vice versa The equilibrium position is determined by ;

Temperature

If the reaction is endothermic, it will be favoured by increase in temperature and equilibrium position will shift to the right ( reactants )

If the reaction is exothermic, its the vice versa

NOTE: Only temperature affects the equilibrium position

5 0

3 years ago