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2 years ago

If 15.6 g of hydrate are heated and only 11.7 g of anhydrous salt remain, calculate the % of water lost.

Chemistry

1 answer:

Elodia [21]2 years ago

8 0

Answer:

Percent loss of water = 25%

Explanation:

Given data:

Mass of hydrated salt = 15.6 g

Mass of anhydrous salt = 11.7 g

Percentage of water lost = ?

Solution:

First of all we will calculate the mass of water in hydrated salt.

Mass of water = Mass of hydrated salt - Mass of anhydrous salt

Mass of water = 15.6 g - 11.7 g

Mass of water = 3.9 g

Now we will calculate the percentage.

Percent loss of water = mass of water / total mass × 100

Percent loss of water = 3.9 g/ 15.6 g × 100

Percent loss of water = 25%

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2 C4H10 + 13 O2 → 8CO2 + 10 H2O Which one is the Limiting Reactant

RUDIKE [14]

Answer:

It's a lot of calculation and it took me a bit of time

...

EXPLANATIONS

Consider the following combustion reaction: 2 C4H10 + 13 O2 ----> 8 CO2 + 10 H2O. 125 g or C4H10 react with 415 g of O2.

a.) Which mass of CO2 and H2O can be produced?

b.) Which substance is the limiting

Molar mass of C4H10,

MM = 4*MM(C) + 10*MM(H)

= 4*12.01 + 10*1.008

= 58.12 g/mol

mass(C4H10)= 125.0 g

use:

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(1.25*10^2 g)/(58.12 g/mol)

= 2.151 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 415.0 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(4.15*10^2 g)/(32 g/mol)

= 12.97 mol

Balanced chemical equation is:

2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O

2 mol of C4H10 reacts with 13 mol of O2

for 2.151 mol of C4H10, 13.98 mol of O2 is required

But we have 12.97 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

According to balanced equation

mol of CO2 formed = (8/13)* moles of O2

= (8/13)*12.97

= 7.981 mol

use:

mass of CO2 = number of mol * molar mass

= 7.981*44.01

= 3.512*10^2 g

According to balanced equation

mol of H2O formed = (10/13)* moles of O2

= (10/13)*12.97

= 9.976 mol

use:

mass of H2O = number of mol * molar mass

= 9.976*18.02

= 1.798*10^2 g

Answer:

mass of CO2 = 3.51*10^2 g

mass of H20 = 1.80*10^2 g

O2 is limiting reagent

3 0

2 years ago

HELP ASAP How many millimeters of mercury pressure does a gas exert at 3.16 atm?

Georgia [21]

The conversion factor is 760 mmHg/atm.

3.1 atm * 760 mmHg/atm = 2356 m

0 0

2 years ago

Read 2 more answers

Rank the size of a change in temperature of one degree Fahrenheit, one degree Celsius, and one kelvin. In other words, if a ther

Lena [83]

The relationship between Fahrenheit and Celsius is:

°F = 1.8*°C + 32

If we differentiate the given expression to find the relationship between a unit change of °F and °C, we get:
Δ°F = 1.8*Δ°C

This means that a change in 1 degree Fahrenheit is equivalent to a change of 1.8 degrees Celsius. Moreover, a one degree change in Fahrenheit is the same as a one-degree change on the Rankine scale, so, ranking the scales in order from higher change to lower change:

1) Fahrenheit = Rankine
2) Celsius = Kelvin

3 0

1 year ago

What is the formula when the halogen from 3rd principal energy level combines with you alkaline earth metal from the 4th energy

Korvikt [17]

Answer: $CaCl_2$

Explanation:

Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....

Halogen with n=3 is chlorine.

Alkaline earth metal with n = 4 is calcium.

An ionic bond is formed when an element completely transfers its valence electron to another element.

Electronic configuration of calcium

$[Ca]=1s^22s^22p^63s^23p^64s^2$

calcium atom will loose one electron to gain noble gas configuration and form calcium cation with +2 charge.

$[Ca^{2+}]=1s^22s^22p^63s^23p^6$

Electronic configuration of chlorine:

$[Cl]=1s^22s^22p^63s^23p^5$

Chlorine atom will gain one electron to gain noble gas configuration and form chloride ion with -1 charge.

$[Cl^-]=1s^22s^22p^63s^23p^6$

Here $Ca^{2+}$ cation and chloride $Cl^{-}$ anion combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $CaCl_2$ .

5 0

2 years ago

A student measures the absorbance of a solution containing FeSCN2 ion using a spectrophotometer. The cuvette used by the student

suter [353]

Answer:

The measured absorbance will be too large.

Explanation:

Fe³⁺(aq) + SCN⁻(aq) ⟶Fe(SCN)²⁺(aq)

A = log₁₀(I₀/I)

If the student orients the cuvette so that the path of the light is through the frosted sides of the cuvette, little light will be able to reach the detector.

The measured intensity (I) will be quite small, so the absorbance (A) will be unusually large.

6 0

2 years ago