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RUDIKE [14]
3 years ago
10

Which chemical agent will you use to modify the frog heart rate?

Chemistry
1 answer:
Leto [7]3 years ago
4 0
There are different chemical agents which are used to study the heart rate of frog. Generally Ringer's solution is used to study frog's heart simulation. At different temperature, frogs heart speeds up or slows down while using Ringer's solution at experiment. Ringer's solution is mixture of salt solution which comprises of NaCl, KCl, CaCl2 and Na2CO3. Sometimes other chemicals like MgCl or antibiotics are also used as addition in Ringer's solution. 
This solution is chiefly used to study in vitro experiments on organs and tissues like frog's heart.

Ringer's solution at 23 degree Celsius for normal heart rate
Ringer's solution at 32 degree Celsius, heart rate speed up and 
Ringer's solution at  5 degree Celsius, heart rate slows down.

other chemicals also have significant effect in heart rate,
For example, Calcium ion in excess will slow down the heart rate, Atropine increases heart rate and digitalis slows down the heart rate.
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Iron (Fe)

Explanation:

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Thallium consists of 29.5%TI-203 and 70.5%TI-205. What is the relative atomic mass of thallium?
VladimirAG [237]
Atomic mass :

29.5 * 203 + 70.5 * 205 / 100 = 

5988.5 + 14452.5 / 100 = 

20441 / 100 = 204.41 u

atomic mass of Thallium <span>204.41 u
</span>
hope this helps!
6 0
3 years ago
Read 2 more answers
A compound contains nitrogen and a metal. This compound goes through a combustion reaction such that compound X is produced from
tia_tia [17]

Answer:

The correct answer is: X is nitrogen dioxide, and Y is a metal oxide

Explanation:

Combustion of compound of containing nitrogen and metal will give nitrogen  dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.

The general equation is given as:

4M_3N_x+7xO_2\rightarrow 4xNO_2+6M_2O_x

Hence, the correct answer is :X is nitrogen dioxide, and Y is a metal oxide.

6 0
4 years ago
Given: Fe2O3(s) + 3 CO(g) –––––&gt; 2 Fe(s) + 3 CO2(g); ΔH° = –26.8 kJ FeO(s) + CO(g) –––––&gt; Fe(s) + CO2(g); ΔH° = –16.5 kJ d
Tpy6a [65]

Answer:

ΔH° = + 6.2 kJ

Explanation:

Fe₂O₃(s) + 3CO(g) –––––> 2 Fe(s) + 3 CO₂(g);        ΔH° = –26.8 kJ  ......... ( 1 )

  FeO(s) + CO(g) –––––> Fe(s) + CO₂(g);                  ΔH° = –16.5 kJ  .........( 2 )

   Multiplying equation ( 2 ) by 2 and subtracting it from ( 1 )

  Fe₂O₃(s) + 3CO(g) -2 FeO(s) -  2CO(g) ––> 2 Fe(s) + 3 CO₂(g) - 2Fe(s) -           2CO₂(g)              ΔH° = –26.8 kJ  -  ( 2 x –16.5 kJ )

Fe₂O₃(s) + CO(g) -2 FeO(s)––>  CO₂(g)              ΔH° = –26.8 kJ + 33 kJ

Fe₂O₃(s) + CO(g) ––>2 FeO(s) +CO₂(g)               ΔH° = + 6.2 kJ

Required   ΔH° = + 6.2 kJ       Ans .

8 0
3 years ago
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