Chất nào dưới đây không phải là este? A. HCOOC6H5 B. HCOOCH3 C. CH3COOH D. CH3COOCH3

Name all 7 elements classified as metalloids. PLEASE HELP ASAP

Strike441 [17]

Answer:

Explanation

Boron (B)

Silicon (Si)

Germanium (Ge)

Arsenic (As)

Antimony (Sb)

Tellurium (Te)

Polonium (Po

6 0

3 years ago

Which of the statements listed below is negative aspect of a volcanic eruption?

mixas84 [53]

Answer:

C . the dramatic scenery created by volcanic eruptions attracts tourists.

5 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

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 $CO_2=1.03g$

Mass of $H_2O=0.14g$

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

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

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 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

People take antacids, such as milk of magnesia, to reduce the discomfort of acid stomach or heartburn. The recommended dose of m

Llana [10]

Answer:

$V_{HCl}=0.208L=208mL$

Explanation:

Hello,

In this case, since the chemical reaction is:

$2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O$

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

$n_{HCl}=2*n_{Mg(OH)_2}$

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

$n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g} =0.00858mol$

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:

$[H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M$

Then, since the concentration and the volume define the moles, we can write:

$[HCl]*V_{HCl}=2*n_{Mg(OH)_2}$

Therefore, the neutralized volume turns out:

$V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL$

Best regards.

3 0

3 years ago

Butanoic acid + 1-pentanol

lutik1710 [3]

Answer: X is an Esther

Explanation: alcohol and carboxylate acid forms esters

7 0

3 years ago