Help me!!! :(<br> Plissss!!!!!

When you graph an inequality you used a closed dot or a open dot

Naddik [55]

For ≤ and ≥ , use a closed dot to indicate the number itself is part of the solution. For < and >, use an open circle to indicate the number itself is not part of the solution.

4 0

3 years ago

I think it’s D but I’m not sure.. Please help!

alisha [4.7K]

Answer: I think it’s D?

Explanation:

7 0

3 years ago

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On increasing which of the following factors, decreases the extent of physisorption? Surface area of the adsorbent Critical temp

Citrus2011 [14]

Answer: On increasing temperature at which adsorption is carried out decreases the extent of physisorption.

Explanation:

An adsorption where molecules of the adsorbate are placed or held on the surface of adsorbent by Vander waals forces is called physisorption.

There is basically physical bonding between the molecules of gas to the surface of a solid or liquid.

Physisorption is reversible in nature and occurs at low temperatures.

It is not specific in nature which means that all gases are adsorbed on the surface of every solid substance to some extent.

Thus, we can conclude that on increasing temperature at which adsorption is carried out decreases the extent of physisorption.

4 0

3 years ago

Define "mole in chemistry terms

Mashutka [201]

Answer:

Can you please post a pictah.

Explanation:

3 0

3 years ago

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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

Help me!!! :( Plissss!!!!!