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3 years ago

13

Ok this is so weird but like how do you expose fake friends for talking about you? I know they did it but i dont know how to tal

k to them.

2 answers:

murzikaleks [220]3 years ago

8 0

Answer:

just do it

Explanation:

Masteriza [31]3 years ago

6 0

Ask them if you can talk in private. DO NOT TEXT THEM (they can screenshot and send to other people). When you get them alone, make sure nobody else is around and then tell them that you know that they’ve been talking about you behind your back and you don’t like that. Tell them how you feel and ask them to stop (I know it’s not always that simple). If they continue, or make fun of you for it, then drop them. Stop talking to them, ignore them, but don’t talk about them behind their back. 2 wrongs don’t make a right

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A boy swings a rubber ball attached to a string over his head in a horizontal, circular path. The piece of string is 1.15 m long

gregori [183]

Answer:

v = 16.49 m/s

Explanation:

Given that,

Length of the string, l = 1.15 m

The ball makes 137 complete turns each minute.

We know that, 1 turn = 6.28 rad

137 turns = 860.79 rad

1 min = 60 s

$\omega=\dfrac{860.79\ rad}{60\ s}\\\\=14.34\ rad/s$

We need to find the tangential velocity of the ball. It can be given by

$v=r\omega\\\\=1.15\times 14.34\\\\v=16.49\ m/s$

So, the tangential velocity of the ball is 16.49 m/s.

5 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

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

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

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

<u>3. Solution</u>

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

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

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

$P_1V_1=P_2V_2$

<u>3. Solution</u>

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

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

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.

<u />

<u>3. Solution</u>

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

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

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

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

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

3 0

3 years ago

What's the difference between a molecular formula and a structural formula

Klio2033 [76]

Glucose and Galactose both have the same molecular formula, C6H12O6, but in the body, galactose must be first converted to glucose to make energy. The difference<span> is their </span>structures

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3 years ago

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33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b

Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water = 18.88 g

Total mass of the solution = Mass of fructose + Mass of water = M

M = 33.56 g + 18.88 g =52.44 g

Volume of the solution = V = 40.00 mL

Density = $\frac{Mass}{Volume}$

a) Density of the solution:

$\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL$

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = $n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol$

Molar mass of water = 18.02 g/mol

Moles of water= $n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol$

Mole fraction of fructose in this solution: $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}$

$\chi_1=0.1510$

Mole fraction of water = $\chi_2=1-\chi_1=0.8490$

c) Average molar mass of of the solution:

= $\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol$

$=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol$

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

$\frac{\text{Average molar mass}}{\text{Density of the mass}}$

$=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol$

5 0

3 years ago

A CO2 fire extinguisher is used for which type of fire?

zlopas [31]

Answer:

Carbon Dioxide fire extinguishers extinguish fire by taking away the oxygen element of the fire triangle and also be removing the heat with a very cold discharge. Carbon dioxide can be used on Class B & C fires. They are usually ineffective on Class A fires

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3 years ago

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