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mezya [45]
3 years ago
11

What is the percentage of water in hydrated calcium chloride​

Chemistry
2 answers:
earnstyle [38]3 years ago
8 0

Answer:

24.5%

Explanation:

You just add up the atomic masses.  

Ca - 40.078  

Cl2 - 35.4527 x 2 = 70.9054  

------ 110.9834  

H4 - 1.00794 x 4 = 4.03176  

O2 - 31.9998  

------ 36.03056  

TOTAL - 147.01396  

So the water is 36.03056/147.01396 = .245082576 but that is only accurate to three decimals (because the mass of Ca was only given to three decimals) so we write .245 and that is 24.5%

This is not my answer but I found it on Yahoo answers and it was answered by Anonymous.

oee [108]3 years ago
8 0

Answer: <u>24.5%</u>

Explanation: Molar mass of CaCl2 * 2 H2O = 40.078 + ( 2 x 35.453) + ( 2 x 18.02)=147.024 g/mol  

% H2O = 2 x 18.02 x 100/ 147.024 =24.5

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garik1379 [7]

Explanation:

vegetable oils have long unsaturated carbon chains which can converted into vegetable ghee

4 0
2 years ago
Calculate the equilibrium number of vacancies per cubic meter for copper at 1000K. The energy for vacancy formation is 0.9eV/ato
nexus9112 [7]

Answer:

Therefore the equilibrium number of vacancies per unit cubic meter =2.34×10²⁴ vacancies/ mole

Explanation:

The equilibrium number of of vacancies is denoted by N_v.

It is depends on

  • total no. of atomic number(N)
  • energy required for vacancy
  • Boltzmann's constant (k)= 8.62×10⁻⁵ev K⁻¹
  • temperature (T).

N_v=Ne^{-\frac{Q_v}{kT} }

To find  equilibrium number of of vacancies we have find N.

N=\frac{N_A\ \rho}{A_{cu}}

Here ρ= 8.45 g/cm³  =8.45 ×10⁶m³

N_A= Avogadro Number = 6.023×10²³

A_{Cu}= 63.5 g/mole

N=\frac{6.023\times 10^{23}\times 8.45\times 10^{6}}{63.5}

   =8.01\times 10^{28 g/mole

Here Q_v=0.9 ev/atom , T= 1000k

Therefore the equilibrium number of vacancies per unit cubic meter,

N_v=( 8.01\times 10^{28}) e^{-(\frac{0.9}{8.62\times10^{-5}\times 1000})

   =2.34×10²⁴ vacancies/ mole

3 0
3 years ago
A long chain of hydrocarbon bonded to cooh is a ________ acid.
vodomira [7]
A long chain of hydrocarbon bonded to COOH is a FATTY acid.
6 0
3 years ago
Read 2 more answers
An industrial manufacturer wants to convert 175 kg of methane into HCN. Calculate the masses of ammonia and molecular oxygen req
Tema [17]
Given:

175 kilograms of Methane (CH4) to be synthesized into Hydrogen Cyanide (HCN)

The balanced chemical equation is shown below:

2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To calculate for the masses of ammonia and oxygen needed, our basis will be 175 kg CH4.

Molar mass:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol

mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol 
mass of NH3 = 185.94 kg NH3 needed

mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
mass of O2 = 525 kg

mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
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4 0
3 years ago
An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at
lana [24]

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

6 0
2 years ago
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