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dimulka [17.4K]
3 years ago
11

An equation that shows only the chemical formulas without

Chemistry
1 answer:
Ivahew [28]3 years ago
7 0

Answer: skeleton equation

Explanation:

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A _____ forecasts the weather using data collected from many sources.
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Meteorologist.............................
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3 years ago
A compound that occupies a receptor but does not activate the neuron is known as a(n)?
Yuliya22 [10]

A compound that binds to a receptor but does not activate the neuron is known as an Antagonist.

A receptor is a large protein molecule on a neuron that gets activated when a ligand binds to it such as a drug or hormone, or when electrical impulses pass through it.

An antagonist is a drug or hormone that binds to receptor, but instead of activating the receptor, it blocks or dampens the activation of the neuron. Antagonist drugs are used to interfere with the normal function or operation of a protein receptor.

Depending on the nature of the antagonist or the receptor it's bound to, the effects of antagonists may be permanent or temporary.

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8 0
1 year ago
A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out
scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0
3 years ago
Helppp pleaseeee......
Cloud [144]

Answer:

mL

Explanation:

milliliters

8 0
3 years ago
Read 2 more answers
If equal volumes of 0.1 M HCl and 0.2 M TRIS (base form) are mixed together. The pKa of TRIS is 8.30. Which of the following sta
blondinia [14]

Answer:

option D is correct

D. This solution is a good buffer.

Explanation:

TRIS (HOCH_{2})_{3}CNH_{2}

if TRIS is react with HCL it will form salt

(HOCH_{2})_{3}CNH_{2} + HCL ⇆   (HOCH_{2})_{3}NH_{3}CL

Let the reference volume is 100

Mole of TRIS is =  100 × 0.2 = 20

Mole of HCL is =  100 × 0.1 = 10

In the reaction all of the HCL will Consumed,10 moles of the salt will form

and 10 mole of TRIS will left

hence , Final product will be salt +TRIS(9 base)

H = Pk_{a} + log (base/ acid)

8.3 + log(10/10)

8.3

6 0
3 years ago
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