3 years ago

Give an example of homeostasis in your body

1 answer:

3 0

Answer:

So, basically Homeostasis is regulation and the. one seen almost every day is shivering or sweating. Our body's normal temperature is approximately 37° C and if you drop below that your body will regulate that by shivering or if you go above it you'll start to sweat to cool off you body and decrease your body temperature back to "equilibrium" or normal temp.

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The circumference of a circle is 31.4 centimeters. What is the circle's radius?<br> Use 3.14 for .

Marizza181 [45]

Radius = 0.5 centimeters

3 0

2 years ago

When 2.3 × 10^3 g of CaCO3 are heated, the actual yield of CaO is 1.09 × 10^3g. What is

iren [92.7K]

The percent yield : 4. 84.58%

<h3>Further explanation</h3>

Reaction

CaCO₃ ⇄ CaO+CO₂

mass CaCO₃ = 2.3 × 10³ g

mol CaCO₃ (MW=100.0869 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{2.3\times 10^3}{100,0869}\\\\mol=22.98$

From the equation, mol CaCO₃ : CaO = 1 : 1, so mol CaO=22.98

mass CaO(MW=56.0774 g/mol)⇒ (theoretical) :

$\tt mass=mol\times MW\\\\mass=22.98\times 56,0774\\\\mass=1288.659~g$

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{1090}{1288.659}\times 100\%\\\\\5yield=84.58\%$

5 0

3 years ago

Which of these statements is an example of a prediction? A;birds are red, blue, and brown. B;birds are large. C;if you walk near

miss Akunina [59]

Answer:

Explanation:

If you walk near them, they will fly away

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3 years ago

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Can anyone help me please

JulijaS [17]

Matter either loses or absorbs energy when it changes from one state to another.

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3 years ago

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Calculate the change in ph when 3.00 ml of 0.100 m hcl(aq) is added to 100.0 ml of a buffer solution that is 0.100 m in nh3(aq)

nikklg [1K]

The change in pH is calculated by:

pOH = Protein kinase B + log [NH4+]/ [NH3]

Protein kinase B of ammonia = 4.74

initial potential of oxygen hydroxide= 4.74 + log 0.100/0.100 = 4.74
pH = 14 - 4.74=9.26

moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100
moles H+ added = 3.00 x 10^-3 L x 0.100 M=0.000300

NH3 + H+ = NH4+
moles NH3 = 0.0100 - 0.000300=0.00970
moles NH4+ = 0.0100 + 0.000300=0.0103

pOH = 4.74 + log 0.0103/ 0.00970= 4.77
oH = 14 - 4.77 = 9.23

the change is = 9.26 - 9.23 =0.03

7 0

3 years ago