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3 years ago

Para que sirven las rondas?? ayuda por favor

Chemistry

1 answer:

prisoha [69]3 years ago

6 0

Answer:

las rondas sirven para ayudar con problemas matematicos asi como cilindros

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The volume of a sample of oxygen is 200.0 mL when the pressure is 3.000 atm and the temperature is 37.0 oC. What is the new temp

attashe74 [19]

Answer:

Explanation:

Given:

V1 = 200 ml

T1 = 20 °C

= 20 + 273

= 293 K

P1 = 3 atm

P2 = 2 atm

V2 = 400 ml

Using ideal gas equation,

P1 × V1/T1 = P2 × V2/T2

T2 = (2 × 400 × 293)/200 × 3

= 234400/600

= 390.67 K

= 390.67 - 273

= 117.67 °C

7 0

3 years 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

Which type of energy uses the photoelectric effect?

ikadub [295]

Answer:

My lovely people the answer is SOLAR

Explanation:

i just know

8 0

2 years ago

An ideal gas sample is confined to 3.0 L and kept at 27 °C. If the temperature is raised to 77 °C and the initial pressure was 1

Nadusha1986 [10]

The gas is confined in 3.0 L container ( rigid container) ⇒ the volume remains constant when the temperature is increased from from 27oC to 77oC and therefore V1=V2 .

<h2>Hope it helps you please mark as brainlist</h2>

6 0

3 years ago

100 ml is drawn from 0.1 M solution of KCl and added to 900 ml of water. What is the

grandymaker [24]

Answer:

The new concentration will be 0.01 M.

Explanation:

To determine the new concentration we use the following formula.

concentration (1) × volume (1) = concentration (2) × volume (2)

concentration (1) = 0.1 M

volume (1) = 100 mL

concentration (2) = unknown

volume (2) = 100 mL + 900 mL = 1000 mL

concentration (2) = [concentration (1) × volume (1)] / volume (2)

concentration (2) = (0.1 × 100) / 1000 = 0.01 M

3 0

3 years ago