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dimulka [17.4K]

3 years ago

11

An equation that shows only the chemical formulas without

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.

Sav [38]

Meteorologist.............................

7 0

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.

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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