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1 year ago

How many moles of hcl are required to make 500 ml of a 3.0 mol/l solution of hcl?

Chemistry

1 answer:

LiRa [457]1 year ago

8 0

Answer:

1.5 moles

Explanation:

To find the number of moles of HCl in 500 mL of a 3 M solution of HCl, we consider moles in 1 liter/ 1000 mL.

3 moles HCl is contained in 1000 mL

x moles is HCl is contained in 500 mL

$= \frac{3 \: \times \: 500}{1000} \: moles \\ = 1.5 \: moles$

Hence the number of moles of HCl in 500 mL is 1.5 moles.

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What is the mass of an atom with 8 protons, 8 electrons, and 8 neutrons?

mafiozo [28]

Answer:An oxygen atom usually has 8 protons, 8 neutrons, and 8 electrons. Looking at the periodic table, oxygen has atomic number 8 and atomic weight 15.999.

Explanation:

5 0

3 years ago

How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2

Leya [2.2K]

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1 mol of KNO₃ gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4 * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

3 0

3 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago

How is mercury obtained and separated from other nearby substances?

VLD [36.1K]

Answer:

Most mercury forms in a sulfide ore called cinnabar, but mercury is also frequently found in small amounts in other ores. A common method for separating mercury from cinnabar is to crush the ore and then heat it in a furnace in order to vaporize the mercury. This vapor is then condensed into liquid mercury form.

Explanation:

7 0

3 years ago

Read 2 more answers

What is the modren system of classificationis based on

Paul [167]

In modern biology, there are three approaches to classifying organisms: systematics, cladistics and molecular evolutionary taxonomy. They are all based on organisms' relation to each other, but use different indicators to assign the degree of relationship

8 0

3 years ago