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3 years ago

How to make 0.01 Molar solution in 30 ml of water

Chemistry

1 answer:

Leviafan [203]3 years ago

4 0

Answer:

You see the concentration decreased ten fold, from 1 M to 0.1 M.

You can simply increase the volume by ten times.

For example

If you have 250 mL of 1 molar HCl, you can add distilled water upto 2500 mL. Now the concentration is 0.1 molar.

If you want to use the serial dilution method, you can use the C1V1=C2V2 equation.

C1 = starting concentration.

C2 = final concentration.

V is for volume.

So let's say you have 500 mL of 1 M HCl and you only want 50 mL of 0.1 M HCl.

1 M x V1 = 0.1 M x 50 mL

V1 = 5 mL

So you take 5 mL of the original (stock) solution and dilute it up to 50 mL. Then you have 50 mL of 0.1 M HCl.

Explanation:

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SnO2 + 2 H2 ——> Sn + 2 H2O

SpyIntel [72]

Answer:

0.15g

Explanation:

Given parameters:

Number of molecules of water = 1.2 x 10²¹ molecules

Unknown:

Mass of SnO₂ = ?

Solution:

To solve this problem, we have to work from the known to the unknown specie;

SnO₂ + 2H₂ → Sn + 2H₂O

Ensure that the equation given is balanced;

Now,

the known species is water;

6.02 x 10²³ molecules of water = 1 mole

1.2 x 10²¹ molecules of water = $\frac{1.2 x 10^{21} }{6.02 x 10^{23} }$ = 0.2 x 10⁻²moles

Number of moles of water = 0.002moles

From the balanced chemical equation:

2 mole of water is produced from 1 mole of SnO₂

0.002 moles of water will be produced from $\frac{0.002}{2}$ = 0.001moles

To find the mass;

Mass = number of moles x molar mass

Molar mass of SnO₂ = 118.7 + 2(16) = 150.7g/mol

Mass = 0.001 x 150.7 = 0.15g

3 0

3 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago

What design element of an astronaut's spacesuits protects him or her from micrometeoroids, which are tiny bits of space debris t

leonid [27]

Answer:

Tough outer shell

Explanation:

Three types of space suits based on their purpose are IVA (Intravehicular activity), EVA (extravehicular activity), and IEVA (intra/extravehicular activity)

Space suits have an outer layer known as the Integrated Thermal Micrometeoroid Garment also known as the TMG or ITMG

The functions of the TMG includes the provision of an insulation medium to the wearer of the suit thereby preventing loss of heat and keeping the wearer warm, protect the wearer from solar radiation which are harmful as well as from Micrometeoroid and debris orbiting in space known as MMOD

The outermost layer of the EMU TMG suits is white Ortho-Fabric that consists of Nomex, Kevlar and Gore-Tex, making it fire resistant, very strong and very tough, and water proof as well as breathable.

3 0

3 years ago

How did you use QPOE?

Jlenok [28]

QPOE Files

The x-ray data are stored in QPOE files (Quick Position-Ordered Events, *.qp) rather than image arrays. These are lists of photons identified by several quantities, including the position on the detector, pulse height, and arrival time. Note that, unlike IRAF images, QPOE files have no associated header file, and are always stored in the current directory, unless explicitly specified otherwise. Non-PROS IRAF tasks can also access QPOE data files in place of image arrays.

8 0

3 years ago

Nvmmmmmmmmmmmmmmmmmmmmmmmmm lol

Vlad1618 [11]

Answer:

coooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooool lol

Explanation:

7 0

3 years ago

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