Metal produce basic of oxide with oxygen true of false

92 grams of ethanol,C2H5OH, is dissolved in 1 liter of solution. Determine the molarity (M)

const2013 [10]

Answer: 1.997 M

Explanation:

molarity = moles of solute/liters of solution or $M = \frac{mol}{L}$

first we have to find our moles of solute (mol), which you can find by dividing the mass of solute by molar mass of solute

mass of solute: 92 g

molar mass of solute: 46.08 g/mol

let's plug it in:

$\frac{92 g}{46.08 g/mol} = 1.997 mol$

next, we plug it into our original equation:

$\frac{1.997mol}{1 L} = 1.997 M$

4 0

3 years ago

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

Pie

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

Now we have to calculate the mass of $HCl$ .

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$

$\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g$

Therefore, the theoretical yield of HCl is, 56.1735 grams

7 0

3 years ago

Al (s) + CuSO4 (aq) → Cu (s) + Al2(SO4)3 (aq)

azamat

Answer:

Explanation:

Al (s) + CuSO4 (aq) → Cu (s) + Al2(SO4)3 (aq)

2Al (s) + 3CuSO4 (aq) → 3Cu (s) + Al2(SO4)3 (aq).

is the balance chemical equation

3 0

4 years ago

Brainliets and a lot of points to ever answers all questions and get all right

Juli2301 [7.4K]

Answer:

try at your own

4 0

3 years ago

Read 2 more answers

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

irinina [24]

Answer:

$m_{KBr}=6.030gKBr$

Explanation:

Hello.

In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

$(T_b-T_b_0)=i*m*K_b$

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

$m=\frac{T_b-T_b_0}{i*K_b}=\frac{(119.3-117.8)\°C}{2*1.48\°C*kg*mol^{-1}}\\ \\m=0.507mol/kg$

Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:

$n_{KBr}=0.507mol/kg*0.1kg=0.0507mol$

Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:

$m_{KBr}=0.0507mol*\frac{119g}{1mol} \\\\m_{KBr}=6.030gKBr$

Best regards.

3 0

4 years ago

Metal produce basic of oxide with oxygen true of false​

Metal produce basic of oxide with oxygen true of false