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alekssr [168]
2 years ago
9

Create a hypothesis that can be tested in an experiment​

Chemistry
1 answer:
sammy [17]2 years ago
8 0
<h3>Please mark me as Brainliest ........</h3>

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How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?
meriva

Answer:

\large \boxed{\text{77.4 mL}}

Explanation:

                Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

    V/mL:     249

c/mol·L⁻¹:  0.0443     0.285

1. Calculate the moles of Ba(OH)₂

\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1  mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}

3. Calculate the volume of HCl

V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}

8 0
3 years ago
How many atoms of iodine are needed to react with one atom of magnesium?
kotegsom [21]

Answer:

uh

Explanation:

4 0
3 years ago
What is the volume of 18.9 g of a liquid that has a density of 0.956 g/ml
Alex_Xolod [135]
The volume is 19.76987448 by taking the known variables mass=18.9g and density=0.956g/ml
To get volume you divide the mass by the density which gives you about
19.77 ml in volume
6 0
3 years ago
(reply with the answer or get reported) Please help and thanks​
alexgriva [62]
1. C) 10
2. C) 2.33 mole
7 0
3 years ago
If 5.12 g of oxygen O2 gas occupies a volume of 6.21L at a certain temperature and pressure, how many grams of oxygen gas will o
ddd [48]

Answer : The mass of O_2 occupy 30.3 L under the same conditions will be, 24.9 grams.

Explanation :

First we have to calculate the moles of O_2

\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}=\frac{5.12g}{32g/mol}=0.16mol

Now we have to calculate the moles of O_2 in 30.3 L by using Avogadro's law.

Avogadro's law : It is defined as the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

V\propto n

or,

\frac{V_1}{n_1}=\frac{V_2}{n_2}

where,

V_1 = initial volume of gas = 6.21 L

V_2 = final volume of gas = 30.3 L

n_1 = initial moles of gas = 0.16 mol

n_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{6.21L}{0.16mol}=\frac{30.3L}{n_2}

n_2=0.781mol

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

Molar mass of O_2 = 32 g/mol

\text{ Mass of }O_2=(0.781moles)\times (32g/mole)=24.9g

Therefore, the mass of O_2 occupy 30.3 L under the same conditions will be, 24.9 grams.

6 0
3 years ago
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