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2 years ago

What is another way that diamonds are useful

1 answer:

7 0

Diamonds are common in the making of jewelry, but they are also used industrially because of their hardness. They are extremely useful in cutting, grinding, and drilling other materials.

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The zinc within a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (s

Mrac [35]

Answer:

Mass of hydrogen gas evolved is 0.0749 grams.

Explanation:

Total pressure of the gases = p = 758 mmHg

Vapor pressure of water = 23.78 mmHg

Pressure of hydrogen gas ,P = p - 23.78 mmHg = 758 mmHg - 23.78 mmHg

P = 734.22 mmHg = $\frac{734.22}{760} atm=0.966 atm$

Temperature of of hydrogen gas ,T= 25°C =298.15 K

Volume of hydrogen gas = V = 0.949 L

Moles of hydrogen gas =n

PV = nRT (Ideal gas equation )

$n=\frac{PV}{RT}=\frac{0.966 atm\times 0.949 L}{0.0821 atm L/mol K\times 298.15 K}$

n = 0.03745 mol

Moles of hydrogen gas = 0.03745 mol

Mass of 0.03745 moles of hydrogen gas = 0.03745 mol × 2 g/mol = 0.0749 g

Mass of hydrogen gas evolved is 0.0749 grams.

3 0

3 years ago

Which is a common way for a scientific calculator to show the number below? <br> 8.1 102

Elodia [21]

Answer: 8.1e+2 or 8.1E+2 (in the scientific notation mode), based on that the number given is 8.1 × 10².

Explanation

Scientific calculators use the letter e or E to show the numbers in scientific notation mode.

These are some examples of numbers that require scientific notation and how they are shown.

Number                 Scientific calculator display

6.022 × 10²³           6.022 e+23
5 × 10 -12              5 E -12

So, understanding that your question is about number 8.1 × 10², the scientific calculator could show 8e+2.

Nevertheless, since 8 .1 × 10² is 810, not a long number, if the display is not in scientific notation mode, it could just display just 810,

7 0

3 years ago

Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

2 years ago

Explain in a short sentence how you can tell a reaction is a decomposition reaction.

mr Goodwill [35]

Answer: A decomposition reaction occurs when one reactant breaks down into two or more products.

Mark me as brainilist pls

3 0

2 years ago

(c) Polar solutes were separated by hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase. How would

Lelu [443]

Answer:

Explanation:

For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.

However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.

5 0

2 years ago