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statuscvo [17]
3 years ago
6

Explain how science can help you make more informed decisions about what to eat.

Chemistry
2 answers:
kvasek [131]3 years ago
6 0
So you wont die from eating too much great food! XD
kotykmax [81]3 years ago
6 0
Well knowing what kind of substances are in the food you eat or even drink can help your body fight illness.
You might be interested in
Which substance is the oxidizing agent in the following reaction? Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O Which substance
aleksandr82 [10.1K]

Answer:

HNO3 is the oxidizing agent

Explanation:

Step 1:

The oxidizing agent is responsible to oxidize another.  but  itself undergoes  a reduction  .

Fe2S3:

Fe has an oxidation number of +3

S has an oxidation number of -2

HNO3:

O has an oxidation number of -2 ( we have 3 times O so this makes -6)

H has an oxidation number of +1

N has an oxidation number of +5

Fe(NO3)3

O has an oxidation number of -2 ( we have 3 times O so this makes -6)

Fe has an oxidation number of +3

Since NO3 has an oxidation number of -1; N has an oxidation number of +5

S alone has an oxidation number of 0

NO2:

O has an oxidation number of -2 (we have 2 times O, this makes -4)

N has an oxidation number of +4

Fe doesn't change from oxidation number. It stays +3

N goes from +5 to +4 → this is a reduction

S goes from -2 to 0 → this is an oxidation

The reducing agent is the compound that contributes the oxidized species (S).

The oxidizing agent contributes the reducing species (N)

The answer is HNO3

5 0
3 years ago
If there is a name/surname you can't make out due to a speaker's manner of speech, what are you supposed to do?
____ [38]

We can call a person by the word gentleman and Sir or from his/her real name.

If there is a name/surname you can't make out due to a speaker's manner of speech then I will call him gentleman or Sir or I will ask him his real name. Gentleman is a word that is used for noble person and Sir word is also used in order to give someone respect.

Call a person with his real name is also comes under the manner of speech so we can conclude that we can call a person by the word gentleman and Sir or from his/her real name.

Learn more about manner of speech here:

Learn more: brainly.com/question/26023566

6 0
3 years ago
Write the balanced molecular, ionic, and net ionic equations for the reaction when a solution of lithium phosphate is mixed with
Kaylis [27]

Answer:

see explaination

Explanation:

Molecular equation;

2Li3PO4(aq) + 3CaCl2(aq) >>>> Ca3(PO4)2(s) + 6LiCl(aq)

Total ionic equation; . Includes all ions ;

6Li^+(aq) + 2PO4^-3(aq) + 3Ca^+2(aq) + 6Cl^-(aq) >>>> Ca3(PO4)2(s) + 6Li^+(aq) + 6Cl^-(aq)

Net ionic equation; remove common ions from total ionic;

2PO4^-3(aq) + 3Ca^+2(aq) >>>> Ca3(PO4)2(s)

6 0
3 years ago
A
BARSIC [14]

The number of C atoms in 0.524 moles of C is 3.15 atoms.

The number of SO_2 molecules in 9.87 moles  SO_2 is 59.43 molecules.

The moles of Fe in 1.40 x 10^{22} atoms of Fe is 0.23 x 10^{-1}

The moles of C_2H_6O in 2.30x10^{24} molecules of C_2H_6O is 3.81.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

A. The number of C atoms in 0.524 mole of C:

6.02214076 × 10^{23} x 0.524 mole

3.155601758 atoms =3.155 atoms

B. The number of SO_2 molecules in 9.87 moles of SO_2:

6.02214076 × 10^{23} x 9.87

59.4385293 molecules= 59.43 molecules

C. The moles of Fe in 1.40 x 10^{22} atoms of Fe:

1.40 x 10^{22} ÷ 6.02214076 × 10^{23}

0.2324754694 x 10^{-1} moles.

0.23 x 10^{-1} moles.

D. The moles of C_2H_6O in 2.30x10^{24} molecules of C_2H_6O:

2.30x10^{24} ÷ 6.02214076 × 10^{23}

3.819239854 moles=3.81 moles

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

6 0
2 years ago
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

ΔH°f(NO(g)) = 148.04 kJ/mol

8 0
3 years ago
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