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3 years ago

Is Sucrose a ionic or covalent bond

Chemistry

1 answer:

yKpoI14uk [10]3 years ago

4 0

Answer:

covalent bond

Explanation:

hope this helps :D

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50 extra points and brainliest if right !!!

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D. Two Plates are colliding.

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What is the formula for manganese (IV) bromide?

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3 years ago

Trial 12 1.Volume of Acid used ( mL) 2020 2.Molarity of acid used (M) 0.0160.016 3.Initial volume reading of base buret 018.6 4.

arsen [322]

Answer:

$V_B = 18.3mL$ -- Volume of base used

$M_B = 0.0175M$ --- Molarity of base

Explanation:

Given

$V_A = 20mL$ -- Volume of acid used

$V_B_1 = 18.6mL$ --- Buret Initial reading

$V_B_2 = 36.9mL$ --- Buret Final reading

$M_A = 0.016M$ --- Molarity of the acid

Solving (a): Volume of base used (VB)

This is calculated by subtracting the initial reading from the final reading of the base buret.

i.e.

$V_B = V_B_2 - V_B_1$

$V_B = 36.9mL - 18.6mL$

$V_B = 18.3mL$

Solving (b): Molarity base (MB)

This is calculated using:

$M_A * V_A = M_B * V_B$

Make MB the subject

$M_B = \frac{M_A * V_A }{V_B}$

This gives:

$M_B = \frac{0.016M *20mL}{18.3mL}$

$M_B = \frac{0.016M *20}{18.3}$

$M_B = \frac{0.32M}{18.3}$

$M_B = 0.0175M$

Solving (c): <em>There is no such thing as average molarity</em>

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3 years ago

find the total pressure of a gas that initially occupied 27 L at 32 degrees Celsius and 2.5 atm, if the final conditions are 12

yawa3891 [41]

Answer: Total pressure of the gas will be 0.716atm.

Explanation: We are given a gas having initial conditions as

V = 27L

T = 32°C = 305K

P = 2.5atm

As the gas remains same, number of moles of a gas will also be same for initial and final conditions. To calculate the number of moles, we use ideal gas equation, which is,

$PV=nRT$ .......(1)

where, R = gas constant = $\text{0.08206 L atm }mol^{-1} K^{-1}$

For calculating number of moles:

$n=\frac{PV}{RT}$

Putting the values of initial condition in this equation, we get

$n=\frac{(2.5atm)(27L)}{\text{(0.08206 L atm }mol^{-1} K^{-1})(305K)}$

n = 2.696 mol

Now, the final conditions are,

V = 88.0L

T = 12°C = 285K

n = 2.696 mol (calculated above)

P = ? atm

Again using equation 1, we get

$P=\frac{nRT}{V}$

$P=\frac{(2.696mol)(\text{0.08206 L atm }mol^{-1} K^{-1})(285K)}{88.0L}$

P = 0.716atm.

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3 years ago

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