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Jet001 [13]
3 years ago
14

How many moles of NaCl must be dissolved in 0.5 L of water to make a 4 mole/L

Chemistry
1 answer:
kvv77 [185]3 years ago
5 0
C = 4 mol/l
v = 0.5 l

n(NaCl)=cv

n(NaCl) = 4 mol/l · 0.5 l = 2 mol

2 moles of NaCl must be dissolved

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When the temperature of an object changes by 10 °C the same temperature change in Kelvins would be
kumpel [21]
The answer to this question would be: <span> 10 °K

Kelvin and Celcius scales are different by 273</span> degrees but their ratio is the same. One degree in Kelvin is equal to one degree in Celcius. That mean,  10 °C change in Celcius would be same as <span> 10 °K changes in Kelvin too. </span>
8 0
3 years ago
Gases A and B are confined to a cylinder and piston and react to form product C. As the reaction occurs, the system loses 1189 J
aliya0001 [1]

Answer:

The change in the internal energy of the system -878 J

Explanation:

Given;

energy lost by the system due to heat, Q = -1189 J (negative because energy was lost by the system)

Work done on the system, W = -311 J (negative because work was done on the system)

change in internal energy of the system, Δ U = ?

First law of thermodynamics states that the change in internal energy of a system (ΔU) equals the net heat transfer into the system (Q) minus the net work done by the system (W).

ΔU = Q - W

ΔU = -1189 - (-311)

ΔU = -1189 + 311

ΔU = -878 J

Therefore, the change in the internal energy of the system -878 J

8 0
3 years ago
A sample of hydrogen gas will behave most like an ideal gas under conditions of?
fenix001 [56]
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6 0
2 years ago
What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as
serious [3.7K]

CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

<span>CaCl2 ---> Ca2+  +  2 Cl-</span>

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

 

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

 

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

Cl- = 0.2 moles / 0.5 moles

<span>Cl- = 0.4 moles             (ANSWER)</span>

6 0
3 years ago
PLEASE HELP ASAP!
choli [55]

Answer:

\large \boxed{\text{D. 710 g}}

Explanation:

1. Calculate the molar mass of Na₂SO₄

\begin{array}{ccc}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 23 & 46\\\text{1S} & 32 & 32\\\text{4O}&16 & 64\\&\text{TOTAL =} & \mathbf{142}\\\end{array}

The molar mass of Na₂SO₄ is 142 g/mol.

2. Calculate the moles of Na₂SO₄

\text{Moles of Na$_{2}$SO}_{4} = \text{2.5 L solution} \times \dfrac{\text{2.0 mol Na$_{2}$SO}_{4}}{\text{1 L solution}} = \text{5.0 mol Na$_{2}$SO}_{4}

3. Calculate the mass of Na₂SO₄

\text{Mass of Na$_{2}$SO}_{4} = \text{5.0 mol Na$_{2}$SO}_{4} \times \dfrac{\text{142 g Na$_{2}$SO}_{4}}{\text{1 mol Na$_{2}$SO}_{4}} = \text{710 g Na$_{2}$SO}_{4}\\\\\text{You need } \large \boxed{\textbf{710 g}} \text{ of Na$_{2}$SO}_{4}

6 0
3 years ago
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