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300.0 mL of a 0.335 M solution of NaI is diluted to 700.0 mL. What is the new concentration of the solution?

Answers

Answer 1

Answer:

Answer: The new concentration of the solution is 0.143 M.

Explanation:

Given: $V_(1)$ = 300.0 mL, $M_(1)$ = 0.335 M

$V_(2)$ = 700.0 mL, $M_(2)$ = ?

Formula used is as follows.

$M_(1)V_(1) = M_(2)V_(2)$

Substitute values into the above formula as follows.

$M_(1)V_(1) = M_(2)V_(2)\n0.335 M * 300.0 mL = M_(2) * 700.0 mL\nM_(2) = 0.143 M$

Thus, we can conclude that the new concentration of the solution is 0.143 M.

Answer 2

Answer:

Final answer:

To find the new concentration of the solution, you can use the formula C1V1 = C2V2. Plugging in the given values, the new concentration of the solution is 0.144 M.

Explanation:

To find the new concentration of the solution, we can use the formula:

C1V1 = C2V2

Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

Plugging in the given values, we get:

(0.335 M)(300.0 mL) = C2(700.0 mL)

Solving for C2, we find the new concentration of the solution to be 0.144 M.

Learn more about concentration of a solution here:

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A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.

Answers

Answer: The molal boiling point elevation constant $k_b$ of X is $2.4^0C/m$

Explanation:

Formula used for Elevation in boiling point :

$\Delta T_b=k_b* m$

or,

$T_b-T^o_b=i* k_b* (w_2* 1000)/(M_2* w_1)$

where,

$T_b-T^o_b =(125.2-120.7)^0C=4.5^0C$

$k_b$ = boiling point constant = ?

m = molality

$w_2$ = mass of solute (urea) = 55.4 g

$w_1$ = mass of solvent X = 500 g

$M_2$ = molar mass of solute (urea) = 60 g/mol

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

$4.5^oC=k_b* (55.4g* 1000)/(60* 500g)$

$k_b=2.4^0C/m$

Thus the molal boiling point elevation constant $k_b$ of X is $2.4^0C/m$

For your sink at home, the flow rate in is 5000 units/hr. Accumulation is 2500 units. What is the accumulation rate if the out flow is 60 units/min? multiple choicea) 2400 units/hr b) 400 units/hr c)100 units/hr d) 1400 units/hr e) 3400 units/hr

f) other ____________________

Answers

Answer:

Option d) 1400 units/hr

Explanation:

Using law of conservation of mass, you can do a mass balance or volume balance:

Volume in - Volume out = Volume accumulation

If you divide by time unit, you get flow balance:

Volume in / time - Volume out / time = Volume accumulation / time

Flow rate in - Flow rate out = Accumulation rate

Here you have:

Flow rate in = 5000 units/h

Flow rate out = 60 units/min

Convert to rate in units/h: (60 units /min) × (60 min/h) = 3600 units/h

Flow rate out = 3600 units/h

Then:

Accumulation rate = 5000 units/h - 3600 units/h = 1400 units/h

What are two functions of the cilia?

Answers

Answer:

- Proper urine flow by signalling the kidney cells.

- They act as mechanoreceptors or sensory receptors.

Explanation:

Motile' (or moving) cilia are found in the lungs, respiratory tract and middle ear. These cilia have a rhythmic waving or beating motion. They work, for instance, to keep the airways clear of mucus and dirt, allowing us to breathe easily and without irritation. They also help propel sperm.
Btw I don’t know if this is what you meant

What is the mass of silver that can be prepared from 1.50 g of copper metal? Cu(s)+2AgNO3(aq)→Cu(NO3)2(aq)+2Ag(s)

Answers

The mass of silver that can be prepared from 1.50 g of copper metal is 5.10 g.

To find the mass of silver that can be prepared from 1.50 g of copper metal, we need to use stoichiometry and the balanced chemical equation.

From the equation, we can see that 1 mole of copper reacts with 2 moles of silver to produce 1 mole of copper(II) nitrate and 2 moles of silver.

We can use the molar masses of copper and silver to convert grams to moles, and then use the mole ratio to find the moles of silver. Finally, we can convert moles of silver back to grams using the molar mass of silver.

Step 1: Convert grams of copper to moles of copper. (1.50 g Cu) / (63.55 g/mol Cu) = 0.0236 mol Cu

Step 2: Use the mole ratio of silver to copper from the chemical equation. (0.0236 mol Cu) × (2 mol Ag / 1 mol Cu) = 0.0473 mol Ag

Step 3: Convert moles of silver to grams of silver. (0.0473 mol Ag) × (107.87 g/mol Ag) = 5.10 g Ag

Molar mass:

Ag = 107.86 g/mol
Cu = 63.54 g/mol

Mole ratio:

Cu(s)+2 AgNO₃(aq)→Cu(NO₃)₂(aq)+2 Ag(s)

63.54 g Cu ------------- 2 x 107.86 g Ag
1.50 g Cu -------------- ??

Mass Ag = 1.50 x 2 x 107.86 / 63.54

Mass Ag = 32358 / 63.54

= 509.25 g of Ag

hope this helps!

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first ignoring ionic strength and activities. a. silver iodate
b. barium sulfate
c. Repeat the above calculations using ionic strength and activities.

Answers

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_(sp) =\text{[Ag$^(+)$][IO$_(3)$$^(-)$]} = s* s = s^(2) = 3.0* 10^(-8)\ns = \sqrt{3.0* 10^(-8)} \text{ mol/L} = 1.7 * 10^(-4) \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_(sp) =\text{[Ba$^(2+)$][SO$_(4)$$^(2-)$]} = (0.02 + s) * s \approx 0.02s = 1.1* 10^(-10)\ns = (1.1* 10^(-10))/(0.02) \text{ mol/L} = 5.5 * 10^(-9) \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = (1)/(2) \sum_(i) {c_(i)z_(i)^(2)}\n\n\mu = (1)/(2) (\text{[Ba$^(2+)$]}\cdot (2+)^(2) + \text{[NO$_(3)$$^(-)$]}*(-1)^(2)) = (1)/(2) (\text{0.02}* 4 + \text{0.04}*1)= (1)/(2) (0.08 + 0.04)\n\n= (1)/(2) *0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(1)^(2)√(0.06) = -0.51* 0.24 = -0.12\n\gamma = 10^(-0.12) = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_(sp) =\text{[Ag$^(+)$]$\gamma_(Ag^(+))$[IO$_(3)$$^(-)$]$\gamma_{IO_(3)^(-)}$} = s*0.75* s * 0.75 =0.56s^(2)= 3.0 * 10^(-8)\ns^(2) = (3.0 * 10^(-8))/(0.56) = 5.3 * 10^(-8)\n\ns =2.3 * 10^(-4)\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(2)^(2)√(0.06) = -0.51*16* 0.24 = -0.50\n\gamma = 10^(-0.50) = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_(sp) =\text{[Ba$^(2+)$]$\gamma_( Ba^(2+))$[SO$_(4)$$^(2-)$]$\gamma_{ SO_(4)^(2-)}$} = (0.02 + s) * 0.32* s* 0.32 \approx 0.02*0.10s\n2.0* 10^(-3)s = 1.1 * 10^(-10)\ns = (1.1* 10^(-10))/(2.0 * 10^(-3)) \text{ mol/L} = 5.5 * 10^(-8) \text{ mol/L}$

What is impossible for a machine to do?A. do a greater amount of work than the amount of work done on the machine
B. apply a force in a direction that is different than the direction of the force applied to the machine
C. move an object a greater distance than the distance that part of the machine was moved
D. apply a force that is less than the force that is applied to the machine

Answers

Answer:

C

Explanation:

move an object a greater distance than the distance that part of the machine was moved

Answer:

a, sorry for late answer!!

Explanation:

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