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4 years ago

How many moles of nacl are required to make 250 ml of a 3.00 m solution?

1 answer:

4 0

The molarity is calculated using the following rule:
molarity = number of moles of solvent / volume of solution (in liters)

We have the volume of solution = 250 ml = 0.25 liters and the molarity = 3 m

Substituting in the equation, we get:
3 = number of moles / 0.25
number of moles = 3 x 0.25 = 0.75 moles

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Which units are used to measure both velocity and speed? Select 3 correct choices.

poizon [28]

Answer:

km/h

mph

iph

Explanation:

I have no idea what mis or dit could be

km/h is kilometers per hour

mph is miles per hour

I assume iph is inches per hour?

4 0

4 years ago

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .

6 0

3 years ago

Consider two different ions. The anion has a valence of -2. The cation has a valence of +2. The two ions are separated by a dist

strojnjashka [21]

Answer:

Force of attraction = 35.96 $\times 10^{27}$ N

Explanation:

Given: charge on anion = -2

Charge on cation = +2

Distance = 1 nm = $10^{-9}$ m

To calculate: Force of attraction.

Solution: The force of attraction is calculated by using equation,

$F = \dfrac{k \times q_1 q_2}{ \r^2}$ ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k = $8.99 \times 10^9 \ Nm^{2}C^{-2}$

F = force of attraction

r = distance between ions.

Substituting all the values in the equation (1) the equation becomes

$F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }$

Force of attraction = 35.96 $\times 10^{27}$ N

6 0

3 years ago

When a parent has alleles for a dominant trait, what traits show up in the offspring?

Effectus [21]

Try putting this in biology category not chemistry so more ppl can help :)

4 0

3 years ago

Which of the following is not a possible termination step in the free radical chlorination of methane? Group of answer choices ∙

vagabundo [1.1K]

Answer: first option is not a termination

∙CH3 + Cl2 → CH3Cl + Cl∙

Explanation:

Since a radical is formed as part of the product it means it's a propagation step and not a termination step, at termination no free radical exist as product

7 0

4 years ago