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2 years ago

Please help! I really don’t understand it

Chemistry

2 answers:

chubhunter [2.5K]2 years ago

7 0

Answer:

C. It makes sure there is a balance inside the body

Explanation:

Hope this helps

Sidana [21]2 years ago

5 0

Answer:

C. It make sure there is a balance inside the body.

Explanation:

Homeostasis in the body tries to maintain the norm for the survival of the body. This includes temperature & amount of fluids inside the body. If they are too out in the end of the spectrum, your brain will tell you that you need to obtain it, such as drinking more water, going to a cool place if you are in the heat, or to obtain warmth if you are in the cold.

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What volume (ml) of fluorine gas is required to react with 1. 28 g of calcium bromide to form calcium fluoride and bromine gas a

Alexandra [31]

144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

<h3>What is Ideal Gas Law ? </h3>

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Presure

V = Volume in liters

n = number of moles of gas

R = Ideal gas constant

T = temperature in Kelvin

Here,

P = 1 atm [At STP]

R = 0.0821 atm.L/mol.K

T = 273 K [At STP]

Now first find the number of moles

F₂ + CaBr₂ → CaF₂ + Br₂

Here 1 mole of F₂ reacts with 1 mole of CaBr₂.

So, 199.89 g CaBr₂ reacts with = 1 mole of F₂

1.28 g of CaBr₂ will react with = n mole of F₂

$n = \frac{1.28\g \times 1\ \text{mole}}{199.89\ g}$

n = 0.0064 mole

Now put the value in above equation we get

PV = nRT

1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K

V = 0.1434 L

V ≈ 144 mL

Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

Learn more about the Ideal Gas here: brainly.com/question/20348074

#SPJ4

4 0

1 year ago

How does electrolysis work

liberstina [14]

Eeletrolysis is a method of removing individual hairs from the face or body .Medical electrolysis devices destroy the growth center of the hair with chemical or heat energy.After a very fine probe is inserted into the hair follicle, the hair is removed with tweezers

5 0

3 years ago

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1. What is the difference between a synthetic resource and a Natural Resource? This is technology.and I need help pls help!!!!!!

Elodia [21]

Answer:

Natural materials are those that are found in nature and have not been made by humans. By comparison, synthetic materials are man-made and cannot be found in nature. Synthetic products are usually created in laboratories by mixing different chemicals, or prepared compounds and substances made in a laboratory.

Explanation:

4 0

3 years ago

Why was Niels Bohr’s atomic model superior to all the earlier models?

gladu [14]

1) it explains about stability of an atom by including stationary state.
2) it explains tge quantization of energy.
3) it gives the concept of angular momentum of a revolving electron.

6 0

3 years ago

Read 2 more answers

How many are molecules ( or formula) in each sample?

andre [41]

Answer:

$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

<u>Explanation</u>:

<u>Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$ </u>

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

<u>Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$ </u>

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

8 0

3 years ago