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2 years ago

A hydrate of CoCl2 with a mass of 6.00 g is heated strongly. after cooling the mass the anhydrate is 3.27g

Chemistry

2 answers:

Oduvanchick [21]2 years ago

7 0

Answer : The amount of water lost from the hydrate is 2.73 grams.

Explanation :

Let us assume that we are calculating the amount of water lost from the hydrate.

As we are given:

Mass of hydrate of $CoCl_2$ = 6.00 g

Mass of anhydrate = 3.27 g

Mass of water lost = Mass of hydrate of $CoCl_2$ - Mass of anhydrate

Mass of water lost = 6.00g - 3.27g

Mass of water lost = 2.73 g

Hence, the amount of water lost from the hydrate is 2.73 grams.

Brums [2.3K]2 years ago

6 0

We are given with a hydrate with a mass of 6 grams and after heating the mass and cooling afterward, the mass went down to 3.27 grams. The amount of water in the hydrate thus is 6-3.27 g or 2.73 grams.That is equivalent to 45.5 percent by mass water.

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A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight

zhannawk [14.2K]

Answer: The molecular formula will be $C_8H_8O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.9}{1.5}=4$

For H = $\frac{5.9}{1.5}=4$

For O = $\frac{1.5}{1.5}=1$

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is $C_4H_4O$

The empirical weight of $C_4H_4O$ = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2$

The molecular formula will be= $2\times C_4H_4O=C_8H_8O_2$

4 0

2 years ago

A 3.5 kg iron shovel is left outside through the winter. The shovel, now orange with rust, is rediscovered in the spring. Its ma

Anton [14]

Answer:

Explanation:

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1 year ago

Lithium acetate, LiCH3CO2, is a salt formed from the neutralization of the weak acid acetic acid, CH3CO2H, with the strong base

Vesna [10]

Answer : The pH of 0.289 M solution of lithium acetate at $25^oC$ is 9.1

Explanation :

First we have to calculate the value of $K_b$ .

As we know that,

$K_a\times K_b=K_w$

where,

$K_a$ = dissociation constant of an acid = $1.8\times 10^{-5}$

$K_b$ = dissociation constant of a base = ?

$K_w$ = dissociation constant of water = $1\times 10^{-14}$

Now put all the given values in the above expression, we get the dissociation constant of a base.

$1.8\times 10^{-5}\times K_b=1\times 10^{-14}$

$K_b=5.5\times 10^{-10}$

Now we have to calculate the concentration of hydroxide ion.

Formula used :

$[OH^-]=(K_b\times C)^{\frac{1}{2}}$

where,

C is the concentration of solution.

Now put all the given values in this formula, we get:

$[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}$

$[OH^-]=1.3\times 10^{-5}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1.3\times 10^{-5})$

$pOH=4.9$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1$

Therefore, the pH of 0.289 M solution of lithium acetate at $25^oC$ is 9.1

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2 years ago

Suppose that on a hot and sticky afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab

uranmaximum [27]

The answer is B) 230 m3

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1 year ago

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How many moles of ions are in 285 ml of 0.0150 m mgcl2?

Solnce55 [7]

MgCl₂)= Mg²⁺ + 2Cl⁻
V(MgCl₂)=285cm³=0,285dm³
c(MgCl₂)=0,015 mol/dm³
n(MgCl₂)=c·V= 0,015 mol/dm³ · 0,285dm³ = 0,0042 mol
n(Mg²⁺)=n(MgCl₂)=0,0042 mol
n(Cl⁻)=2n(MgCl₂)=0,0084 mol

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2 years ago

Read 2 more answers