What is the correct noble gas configuration for sr?

The end point of a titration was reached after 22 ml of 0.050 m disodium edta titrant was dispensed into a solution containing t

ale4655 [162]

The moles of disodium edta used is calculated using the below formula

moles =molarity x volume in liters

molarity=0.050m
volume in liters = 22/1000=0.022 L

moles is therefore= 0.022 x0.050 =1.1 x10^-3 moles of disodium edta

4 0

3 years ago

Read 2 more answers

When filtered through a funnel into a flask, a mixture of substances X, Y and Z gets separated as below: - X stays in the funnel

alisha [4.7K]

Z is the solvent, Y is soluble in water while X is insoluble in water.

<h3>Filtration</h3>

Filtration is a method of separation of substances based on particle size. Only a particular particle size can pass through the filter. The substance that remains in the filter is the residue while the substances that passes through the filter is called the filtrate.

From the observation in the question Z is the solvent, Y is soluble in water while X is insoluble in water.

Learn more about separation of mixtures: brainly.com/question/863988

6 0

2 years ago

Lucille is finding it difficult to play soccer after school. Her doctor thinks that her cells might not be getting enough oxygen

Natali5045456 [20]

Answer:

They might not be bringing in enough oxygen from the air. It's hard to play soccer if you don't have enough oxygen in your cells. They might not be breaking down starch to make glucose.

Explanation:

6 0

2 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

2 years ago

A graduated cylinder contains 18.0ml of water. what is the new water level after 35.6g of silver metal with a density of 10.5 g/

AlekseyPX

Answer:

The answer to your question is

1.- Volume = 3.4 ml

2.- Volume = 0.61 ml

3.- Mass = 2872.8 pounds

Explanation:

Problem 1

Volume = 18 ml

mass = 35.6 g

density = 10.5 g/ml

Process

1.- Calculate the volume of silver

Formula

$density = \frac{mass}{volume}$

solve for volume

$volume = \frac{mass}{density}$

Substitution

$volume = \frac{35.6}{10.5}$

volume = 3.4 ml

2.- Problem 2

Total volume = ?

Volume = 18 + 3.4

Volume = 21.4 ml

Data

mass = 8.3 g

density = 13.6 g(ml

volume = ?

Formula

$density = \frac{mass}{volume}$

Solve for volume

$volume = \frac{mass}{density}$

Substitution

$volume = \frac{8.3}{13.6}$

Result

volume = 0.61 ml

3.- Problem 3

Data

volume = 345 gal

density = 1 g/ml

mass = ?

Formula

$density = \frac{mass}{volume}$

Solve for mass

mass = density x volume

Covert gal to ml

1 gal --------------- 3785 ml

345 gal ------------- x

x = (345 x 3785) / 1

x = 1305825 ml

Substitution

mass = 1 x 1305825

mass = 1305825 g

Convert g to pounds

1 g ------------------- 0.0022 pounds

1305825 g ---------------- x

x = (1305825 x 0.0022)

x = 2872.8 pounds

5 0

2 years ago