All categories

2 years ago

Achieving a balanced diet is best done by

1 answer:

8 0

B. eating the right amount of calories each day
for a, you need proteins,veggies and calories to survive, as each does a specific job such as glucose and protein being created, etc.
For c, you can have at least 1 item from each food group each day, but you can still have too much of something and thats harmful.
For d,getting 100% calcium only works with calcium, and you can do anything with the other foods.
For b(which is correct), people usually eat an excess of calories, and that mainly controls your weight. Now other food groups can also affect you if you have too much, but people tend to have too much calories, so getting the right amount of calories is the best choice

You might be interested in

what is 19.8-17.1

Ostrovityanka [42]

Answer: it should be 2.8

Explanation:

7 0

2 years ago

Read 2 more answers

I think it’s b but I’m not sure

suter [353]

Yes that’s correct it is B

4 0

2 years ago

Read 2 more answers

The formation constant for the reaction ag (aq) 2nh3(aq) ag(nh3)2 (aq) is kf = 1.7 × 107 at 25°c. what is δg° at this temperatur

anzhelika [568]

The value of ΔG° at this temperature is -18034.18 J/mol

Calculation,

Given information

formation constant (Kf)= 1.7 × $10^{7}$

Universal gas constant (R) = 8.314 J/K• mol

Temperature = 25° C = 25 °C + 273 = 300 K

Formula used:

ΔG° = -RT㏑Kf

By putting the valur of R,T, Kf we get the value of ΔG°

ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 × $10^{7}$

ΔG° = -2494.2㏑ 1.7 × $10^{7}$ = -18034.18 J/mol

So, change in standard Gibbs's free energy is -18034.18 J/mol

Learn about formation constant

brainly.com/question/14011682

#SPJ4

8 0

1 year ago

Find the number of kilograms in 15.40 moles of C5H10O5

OLga [1]

C5H10O5 would weight a total of 2.312 kilograms.

5 0

2 years ago

The expected o-c-h angle in this molecule is degrees. the expected hybridization at the central carbon is

WINSTONCH [101]

Each neutral carbon atom contains four valence electrons and may form up to four electron domains. Possible hybridizations include

$sp^{3}$ , four electron domains, as in ethane $\text{C}_2\text{H}_6$
$sp^{2}$ , three electron domains, as in ethene $\text{C}_2\text{H}_4$
$sp$ , two electron domains, as in ethyne $\text{C}_2\text{H}_2$

Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

Carbon atoms with a $sp^{3}$ hybridization would demonstrate a tetrahedral configuration with a bond angle of approximately $109.5\textdegree{}$
Carbon atoms with a $sp^{2}$ hybridization would demonstrate a triangular planar configuration with a bond angle of $120\textdegree{}$
Carbon atoms with a $sp$ hybridization would demonstrate a linear configuration with a bond angle of $180\textdegree{}$

Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible

$sp^{3}$ in which the oxygen atom forms a carbon-oxygen double bond with the central carbon atom;
$sp^{2}$ in which the oxygen atom forms a single bond with the central carbon atom and with a second atom.

3 0

3 years ago