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2 years ago

How do people start alcohol consumption

Chemistry

1 answer:

Debora [2.8K]2 years ago

6 0

Answer:

Explanation:Due to the mental pressure,

Due to peer pressure,

Lack of love and affection from the family members.

Influence from the T.V advertisement.

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What type of bond is expected between two carbon atoms?

tatuchka [14]

<h3>Answer:</h3><h2>Covalent Bonds</h2>

<em><u>Carbon Forms Covalent Bonds</u></em>

<em><u>Carbon Forms Covalent BondsThe most common type of bond formed by carbon is a covalent bond. In most cases, carbon shares electrons with other atoms (usual valence of 4). This is because carbon typically bonds with elements which have a similar electronegativity.Jul 28, 2019</u></em>

Explanation:

<h3><em>Hope</em><em> it</em><em> works</em><em> out</em><em>!</em></h3>

3 0

2 years ago

A 1.775g sample mixture of potassium hydrogen carbonate is decomposed by heating. if the mass loss is 0.275g what is the percent

Marina86 [1]

A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.

<h3>What is a decomposition reaction?</h3>

A decomposition reaction can be defined as a chemical reaction in which one reactant breaks down into two or more products.

Step 1: Write the balanced equation for the decomposition of KHCO₃.

2 KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l)

The mass loss of 0.275 g is due to the gaseous CO₂ that escapes the sample.

Step 2: Calculate the mass of KHCO₃ that formed 0.275 g of CO₂.

In the balanced equation, the mass ratio of KHCO₃ to CO₂ is 200.24:44.01.

0.275 g CO₂ × 200.24 g KHCO₃/44.01 g CO₂ = 1.25 g KHCO₃

Step 3: Calculate the mass percentage of KHCO₃ in the sample.

There are 1.25 g of KHCO₃ in the 1.775 g sample.

%KHCO₃ = 1.25 g/1.775 g × 100% = 70.4%

A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.

Learn more about decomposition reactions here: brainly.com/question/14219426

7 0

2 years ago

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

2 years ago

What is the mass of H2O produced when 14.0 grams of H2 reacts completely with 2.0 grams of O2?

Vladimir79 [104]

Balanced chemical equation:

2 H2 + 1 O2 = 2 H2O

4 g H2 -------> 32 g O2 -----------> 36 g H2O
↓ ↓ ↓
14.0 g ---------> 2.0 g O2 ----------> mass H2O ?

32 * mass H2O = 2.0 * 36

32 * mass H2O = 72

mass of H2O = 72 / 32

mass of H2O = 2.25 g

hope this helps!.

4 0

2 years ago

What is the volume of 13.21 g of ethane gas (C2H6) at STP?

yuradex [85]

The volume at STP : 9.856 L

<h3>Further explanation</h3>

Given

Mass of ethane : 13.21 g

Required

The volume at STP

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

mol ethane(C2H6) :

= mass : molar mass

= 13.21 g : 30 g/mol

= 0.44

Volume at STP :

= 0.44 x 22.4 L

= 9.856 L

6 0

2 years ago

How do people start alcohol consumption​

How do people start alcohol consumption