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Fed [463]
2 years ago
13

A 5.0 g sample of calcium nitrate (Ca(NO3)2, M = 164) contaminated with silica (SiO2, M = 60.1) is found to contain 1.0 g calciu

m. What is the mass percent purity of calcium nitrate in the sample?
Chemistry
1 answer:
zhenek [66]2 years ago
4 0

Answer:

= 82%

Explanation:

Percentage purity is calculated by the formula;

% purity = (mass of pure chemical/total mass of sample) × 100

In this case;

1 mole of Ca(NO3)2 = 164 g

but; 164 g of Ca(NO3)2 = 40 g Ca

Therefore; mass of Ca(NO3)2 = 164 /40

                                                  = 4.1 g

Thus;

% purity of Ca(NO3)2 = (Mass of Ca(NO3)2/ mass of the sample)× 100

                                    = (4.1 g/ 5 g) × 100

                                    = 82%

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Water at 4.4°C is to flow through a horizontal, commercial steel pipe having a length of 305 m at the rate of 150 gal/min. If a
Pavel [41]

Answer:

d = 70.5 mm

Explanation:

given,

length of pipe = 305 m

discharge rate = 150 gal/min

pipe diameter = ?

1 gal/min = 6.30902 ×  10⁻⁵ m³/s

150 gal/min = 150 × 6.30902 ×  10⁻⁵ m³/s

                   = 9.46 × 10⁻³ m³/s

h = \dfrac{flv^2}{2gd}

h = \dfrac{flv^2}{2gd}

Q = A V

h = \dfrac{fl(\dfrac{Q}{A})^2}{2gd}

h = \dfrac{fl(\dfrac{Q}{\dfrac{\pi}{4}d^2})^2}{2gd}

h= \dfrac{8flQ^2}{\pi^2gd^5}

f = 0.048 from moody chart using P/D = 0.00015

\dfrac{1}{d^5}= \dfrac{6.1\times \pi^2\times 9.8}{8\times 0.048\times 305\times 0.00946^2}

d = 70.5 mm

Diameter of the pipe is equal to 70.5 mm

7 0
2 years ago
Which statement is correct? Salts are formed by the reaction of bases with water. Most salts are ionic and are soluble in water.
Studentka2010 [4]
<span>Salts are formed by the reaction of bases with water. -  FALSE
</span><span>Most salts are ionic and are soluble in water. -  TRUE
</span><span>Most salts are insoluble in water and lack electrical charges. - FALSE
</span><span>Solutions of salt and water do not conduct electricity. - FALSE

:)</span>
3 0
2 years ago
Read 2 more answers
Oxygen at 150 k and 41.64 atm has a tabulated specific volume of 4.684 cm3/g and a specific internal energy of 1706 j/mol. calcu
docker41 [41]

The total energy in a system due to the temperature and pressure per unit mass in that system is known as specific enthalpy. It is used in thermodynamic equations when one desires to know the energy for a given single unit mass of a component.  

Specific enthalpy is calculated by the equation:  

H = U + PV

in the given case, Specific volume = 4.684 cm³/g = 149.888 cm³/g moles = 149.888 × 10⁻³ J/g moles

Specific internal energy (U) is 1706 J/mol and pressure is 41.64.  

H = 1706 + 41.64 × 149.888 × 10⁻³ × 101.3 joules  

= 2428 joules / mole


5 0
2 years ago
A gas is initially at a pressure of 0.43 atm, and a volume of 11.7 liters. Then the pressure is raised to 3.61 atm and the volum
lukranit [14]

Answer:

D. 91.98K

Explanation:

The General Gas Law equation is given by,

\frac{P_1V_1}{T_1}  =  \frac{P_2V_2}{T_2}

From the question,

the initial pressure,

P_1 = 0.43 \: atm

the initial volume,

V_1 = 11.7 \: litres

the final temperature,

T_2=627K

the final pressure,

P_2=3.61atm

the final volume,

V_2=9.5litres

Making

T_1

the subject of the expression, we obtain

T_1= \frac{P_1V_1T_2}{P_2V_2}

By substitution,

T_1= \frac{0.43 \times11.7 \times627}{3.61 \times9.5}

T_1=91.980K

Hence the initial temperature was 91.98 K

4 0
2 years ago
Read 2 more answers
A student performed an analysis of a sample for its calcium content and obtained the following results:
sweet-ann [11.9K]

Answer:

14.9075 g, 28.67%, 0.11%

Explanation:

The mean concentration of calcium = summation x / frequency

= ( 14.92 + 1491 + 14.88 + 14.92 ) /4 = 14.9075 g

Standard deviation = √(summation (x - μ)² /n) = √ ( ((14.92 - 14.9075)² +(14.91 - 14.9075)² + (14.88 - 14.9075)² + ( 14.92 - 14.9075)²) / 4) = 0.0164

b)  percent error = abs(14.9075 - 20.90) / 20.90 × 100 = 28.67%

c) relative standard deviation = standard deviation / mean × 100 = 0.0164 /  14.9075 × 100 = 0.11%

d) The accuracy of the measure is the measurement compared to the actual which according to the standard set by the instructor (5%error) is not very accurate because the percent error is high (28.67%) while the relative standard deviation is quite low ( 0.11%) which means the measurement precision is very high.

The student will have to redo the experiment because the experiment was not too accurate since the percent error is way higher than the set value (5%) although the precision was high.

5 0
2 years ago
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