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2 years ago

How many grams of solute are dissolved in 125.0 mL of 5.00 M NaCl (MM = 58.45)?

Chemistry

1 answer:

Nookie1986 [14]2 years ago

5 0

Answer: 36.53g

Explanation:

First we need to find the amount of NaCl that dissolves in 1L of the solution that produced 5M of NaCl

Molarity = 5M

MM of NaCl = 58.45

Molarity = Mass conc (g/L) / MM

Mass conc. (g/L) of NaCl = Molarity x MM

= 5 x 58.45 = 292.25g

Next, we need to find the amount that will dissolve in 125mL(i.e 0.125L)

From the calculations above,

292.25g of NaCl dissolved in 1L

Therefore Xg of NaCl will dissolve in 0.125L of the solution i.e

Xg of NaCl = 292.25 x 0.125 = 36.53g.

Therefore 36.53g of NaCl will dissolve in 125mL of the solution

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How many molecules are in 3.6 grams of NaCl?

raketka [301]

Answer:

$\boxed{\text{None}}$

Explanation:

There are no molecules in NaCl, because it consists only of ions.

However, we can calculate the number of formula units (FU) of NaCl.

Step 1. Calculate the moles of NaCl

$\text{No. of moles}=\text{3.6 g NaCl}\times \dfrac{\text{1 mol NaCl}}{\text{63.54 g NaCl}} = \text{0.0567 mol NaCl}$

Step 2. Convert moles to formula units

$\text{FU} = \text{0.0567 mol NaCl} \times \dfrac{6.022 \times 10^{23}\text{ FU NaCl}}{\text{1 mol NaCl}}\\\\= 3.4 \times 10^{22} \text{ FU NaCl}$

There are $\boxed{3.4 \times 10^{22} \text{ FU NaCl}}$ in 3.6 g of NaCl.

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2 years ago

How many moles NaOH are contained in 65.0 mL of a 2.20 M solution of NaOH in H2O?

saw5 [17]

The number of moles of NaOh that are contained in 65ml of 2.20M solution NaOh in H2o is calculated using the below formula

moles = molarity x volume /1000

that is 65 x2.20 /1000= 0.143 moles

0 0

2 years ago

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What is the pH of a solution of 0.400 M CH₃NH₂ containing 0.250 M CH₃NH₃I? (Kb of CH₃NH₂ is 4.4 × 10⁻⁴)

Karolina [17]

Answer:

$\boxed{\text{10.84}}$

Explanation:

A solution of a weak base and its conjugate acid is a buffer.

The equation for the equilibrium is

$\rm CH$_3$NH$_2$ + H$_2$O $\, \rightleftharpoons \,$ CH$_3$NH$_2$+ H$_{3}$O$^{+}$\\\text{For ease of typing, let's rewrite this equation as}\\\rm B + H$_2$O $\longrightarrow \,$ BH$^{+}$ + OH$^{-}$; $K_{\text{b}}$ = 4.4 \times 10^{-4}$$

The Henderson-Hasselbalch equation for a basic buffer is

$\text{pOH} = \text{p}K_{\text{b}} + \log\dfrac{[\text{BH}^{+}]}{\text{[B]}}$

Data:

[B] = 0.400 mol·L⁻¹

[BH⁺] = 0.250 mol·L⁻¹

Kb = 4.4 × 10⁻⁴

Calculations:

(a) Calculate pKb

pKb = -log(4.4× 10⁻⁴) = 3.36

(b) Calculate the pH

$\text{pOH} = 3.36 + \log \dfrac{0.250}{0.400} = 3.36 + \log 0.625 = 3.36 - 0.204 = 3.16\\\\\text{pH} =14.00 -3.16 = \mathbf{10.84}\\\\\text{The pH of the solution is }\boxed{\textbf{10.84}}$

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2 years ago

3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con

KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

$m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}$

c) Half life for first-order kinetics is computed by:

$t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}} =53.3min$

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

$C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M$

Best regards.

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2 years ago

A mixture of KCl and KNO3

Tems11 [23]

Answer is: <span>The percent of KCl in the mixture is closest to </span><span>a) 40%.
</span>
If we use 100 grams of mixture:
ω(K) = 44.20% ÷ 100% = 0.442.
m(K) = 0.442 · 100 g = 44.2 g.
n(K) = 44.2 g ÷ 39.1 g/mol.
n(K) = 1.13 mol.
n(KCl) + n(KNO₃) = n(K)
m(KCl) = x.
m(KNO₃) = y.
Two equtions:
1) x + y = 100 g.
2) m(KCl)/M(KCl) + m(KNO₃)/M(KNO₃) = 1.13 mol.
x/74.55 g/mol + y/101.1 g/mol = 1.13 mol.
From first equation find x = 100 - y and put in second equation:
(100 - y)/74.55 + y/101.1 = 1.13 /×101.1.
135.61 - 1.356y + y = 114.24.
0.356y = 22.37 g.
y = 62.83 g.
x = m(KCl) = 100 g - 62.83 g = 37.17 g.

8 0

2 years ago

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