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3 years ago

Why has diamond has a high melting point

1 answer:

6 0

Well since covalent bonds are strong and diamonds contains a lot of covalent bonds, it makes the diamond's melting point and boiling point very high.

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The book procedure calls for 2,5-dimethylhexane-2,5-diol and concentrated hydrochloric acid to be shaken in a reaction vial. Whi

san4es73 [151]

Answer:

E) All of the above.

Explanation:

Hello,

Since the acidic nature of the HCl implies its corrosiveness, when it is in contact with the skin and eyes the burning starts immediately, so gloves and goggles must be worn. Next, the fuming hydrochloric acid (37% by mass) is volatile so it gives off even when dissolved into water, so it must be used in the fume hood. Then, since vapors are produced during the chemical reaction, an overpressure could be attained, that's why we must keep the glass sash of the fume hood between us and the vial. As a common risk, the vial could be dropped causing the hydrochloric acid to splash, so we must keep the vial well inside the hood.

Best regards.

7 0

3 years ago

21 mL It required 42.35 mL of H2SO4 to neutralize 21.17 mL of 0.5000 M NaOH. Calculate the concentration of H2SO4

bija089 [108]

The balanced equation for the above reaction is
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of NaOH moles required-0.5000 M / 1000 mL/L x 21.17 mL = 0.010585 mol
According to stoichiometry, acid moles required are 1/2 of the base moles reacted
Therefore number of H₂SO₄ moles reacted - 0.010585 /2 mol
Number of moles in 42.35 mL of H₂SO₄ - 0.010585 /2 mol
Therefore in 1 L solution - (0.010585) /2 / 42.35 mL x 1000 mL/L = 0.125 M
Molarity of H₂SO₄ - 0.125 M

5 0

3 years ago

GUYS I REALLY NEED THIS ASAP!!!!!

Nastasia [14]

Answer:

I don't think so

Explanation:

The equation doesn't look balanced

8 0

3 years ago

PLS HELP-CHEMISTRY: What is the molar concentration of an ammonia solution in which there are 0.1557 grams

harina [27]

Answer: M = 0.036 M

Explanation: Solution attached:

First convert mass of NH3 to moles

Next convert volume in mL to L

Use the formula for Molarity

M = n / L

4 0

3 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago