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2 years ago

3.50 grams of an unknown metal was placed in a graduate cylinder containing 50.0

1 answer:

3 0

For 3.50 grams of an unknown metal was placed in a graduate cylinder containing 50.0mL water initially, the density of this unknown meta is mathematically given as

D= 0.23 g/ml

<h3>What is the density of this unknown metal if the final reading from the graduate cylinder was 65.0 mL?</h3>

Generally, the equation for the Volume of water is mathematically given as

Vw= Volume of water after metal addition - Volume of water initial

Therefore

Vw = 65 ml - 50 ml

Vw= 15 ml

In conclusion,

D= Mass of metal / Volume

D= 3.5 / 15

D= 0.23 g/ml

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How do scientists prevent the nonpolar perfluorocarbons

Pavel [41]

Scientists prevent the nonpolar perfluorocarbons

in Oxygent from separating from the water by adding surfactant.

<u>Explanation:</u>

The collection of oxygen from lungs and transporting it all those tissues present in the human body and taking carbon dioxide back to lungs are the main functions of the hemoglobin. But, instead of this hemoglobin, When perfluorocarbons , a nonpolar chemical is used to perform the same function for the oxygen transportation, the nonpolar chemical may not mix well with the water when this perfluorocarbons is taken through a water-based saline .

For this purpose an addition of a chemical called surfactant is done. This will enhance the chemical bonding between the non polar chemical and water. The perfluorocarbons spreads into some droplets and gets coated with this bonding substance. Thus it gets attached to the water.

3 0

3 years ago

Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given

professor190 [17]

Answer: -227 kJ

Explanation:

The balanced chemical reaction is,

$C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]$

$-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]$

$\Delta H_{C_2H_2}=-227kJ$

Therefore, the enthalpy change for $C_2H_2$ is -227 kJ.

7 0

2 years ago

Express the concentration of a 0.0390 M aqueous solution of fluoride, F − , in mass percentage and in parts per million (ppm). A

Vadim26 [7]

Answer:

Mass percentage → 0.074 %

[F⁻] = 741 ppm

Explanation:

Aqueous solution of flouride → [F⁻] = 0.0390 M

It means that in 1L of solution, we have 0.0390 moles of F⁻

We need the mass of solution and the mass of 0.0390 moles of F⁻

Mass of solution can be determined by density:

1g/mL = Mass of solution / 1000 mL

Note: 1L = 1000mL

Mass of solution: 1000 g

Moles of F⁻ → 0.0390 moles . 19g /1 mol = 0.741 g

Mass percentage → (Mass of solute / Mass of solution) . 100

(0.741 g / 1000 g) . 100 = 0.074 %

Ppm = mass of solute . 10⁶ / mass of solution (mg/kg)

0.741 g . 1000 mg/1g = 741 mg

1000 g . 1 kg/1000 g = 1kg

741 mg/1kg = 741 ppm

5 0

2 years ago

Where would you have the least mass?

beks73 [17]

Answer:

the moon but mass stays the same

Explanation:

7 0

2 years ago

Which property among the following is best suitable for metalloids

QveST [7]

Answer:

2. Have the property of metals and non-metals

Explanation:

A metalloid is a substance that posseses the characteristics of both metals and non-metals. In other words, it can be said to be an intermediary between metals and nonmetals.

According to this question, a metalloid

4 0

2 years ago