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4 years ago

How many aluminum atoms are in 3.28 g of al2s3?

1 answer:

6 0

m Al₂S₃: 27g×2 + 32g×3 = 150 g/mol

150 g/mol Al₂S₃ ==>> 2 moles of Al ==>> 12,04·10²³ atoms of Al
--------------------------------------
150g ------- 12,04·10²³ Al
3,28g ------ X
X = (3,28×12,04·10²³)/150g
X = 0,2633·10²³ = 2,633·10²² aluminium atoms

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Calculate the molecular weight of a substance. In which the solution of this substance in the water has a concentration of 7 per

Eduardwww [97]

Answer : The molecular weight of a substance is 157.3 g/mol

Explanation :

As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.

Mass of solute = 7 g

Mass of solution = 100 g

Mass of solvent = 100 - 7 = 93 g

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=k_f\times\frac{\text{Mass of substance(solute)}\times 1000}{\text{Molar mass of substance(solute)}\times \text{Mass of water(solvent)}}$

where,

$\Delta T_f$ = change in freezing point

$T_f^o$ = temperature of pure water = $0^oC$

$T_f$ = temperature of solution = $-0.89^oC$

$K_f$ = freezing point constant of water = $1.86^oC/m$

m = molality

Now put all the given values in this formula, we get

$(0-(-0.89))^oC=1.86^oC/m\times \frac{7g\times 1000}{\text{Molar mass of substance(solute)}\times 93g}$

$\text{Molar mass of substance(solute)}=157.3g/mol$

Therefore, the molecular weight of a substance is 157.3 g/mol

7 0

4 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

3 0

3 years ago

PLEASEEE HELP !!! How many moles of H2O are equivalent to 97.3 grams of H2O? ___? = mol

Aleks04 [339]

no. of moles=mass/molar mass

=97.3/18

=5.40555555556 moles

5 0

4 years ago

2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

8 0

2 years ago

Which would be considered a substance? Element, Compound, Mixture.

Romashka [77]

Answer:

MIXTURE , ELEMENT AND SUBSTANCE

3 0

3 years ago