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3 years ago

Which mass measurement contains four significant figures?

Chemistry

2 answers:

Snezhnost [94]3 years ago

7 0

Answer:

the answer is :

C) 1003 g

egoroff_w [7]3 years ago

7 0

Answer:

Explanation:

i looked up the answer trust me!!!

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22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass

Setler [38]

Answer:

9.82 g of Mg(NO₃)₂

Explanation:

Let's determine the reaction:

2AgNO₃ + MgBr₂ → Mg(NO₃)₂ + 2AgBr

2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

We determine the moles of AgNO₃

22.5 g . 1mol / 169.87g = 0.132 moles

Ratio is 2:1.

2 moles of silver nitrate can produce 1 mol of magnesium nitrate

Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles

We convert moles to mass:

0.0662 mol . 148.3 g/ mol = 9.82 g

6 0

3 years ago

How are planes contrail lines similar to clouds

Tom [10]

Both are mainly composed of droplets of condensed water

8 0

3 years ago

The spectrochemical series is I < Br< < Cl^- < F

vfiekz [6]

Answer:

b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

Explanation:

The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.

Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.

Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.

Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

6 0

3 years ago

Question 4

nata0808 [166]

Answer:

A. She can run the same distance in a shorter amount of time.

Explanation:

Speed can be computed with the formula:

$v = \dfrac{d}{t}$

Where:

v = speed

d = distance

t = time

So if you look at the formula, distance there tells you how far an object has traveled, and the time tells us how long it took to cover that distance. As you can see, speed and time are <u>inversely proportional. </u>This means that as one goes up, the other goes down. So if we reduce time travelling a certain distance, the speed increases.

3 0

3 years ago

The average C - H bond energy in CH4 is 415 kJ / mol. Use the following data to calculate the average C - H bond energy in ethan

sertanlavr [38]

Answer and Explanation:

ΔH = (Bond energies of the products) - (Bond Energies of the reactants)

a) C2H6 + H2 ----> 2CH4 ΔH = -65.07 KJ/mol

Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol

Bond energy of reactants = Bond energy of C2H6 + Bond energy of H2

Bond energy of C2H6 = (C-C) + 6(C-H) = 347 + 6(C-H)

Bond energy of H2 = (H-H) = 432 KJ/mol

-65.07 = 3320 - (Bond energy of C2H6 + 432)

Bond energy of C2H6 = 3320-432+65.07 = 2953.07 KJ/mol

Bond energy of C2H6 = (C-C) + 6(C-H) = 347 + 6(C-H) = 2953.07

(C-H) = (2953.07 -347)/6 = 434.345 KJ/mol

b) C2H4 + 2H2 ------> CH4 ΔH = -202.21 KJ/mol

Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol

Bond energy of reactants = Bond energy of C2H4 + (2 × Bond energy of H2)

Bond energy of C2H4 = (C=C) + 4(C-H) = 614 + 4(C-H)

(2 × Bond energy of H2) = 2 × 432 = 864 KJ/mol

-202.21 = 3320 - (Bond energy of C2H4 + 864)

Bond energy of C2H4 = 3320+202.21-864 = 2658.1 KJ/mol

Bond energy of C2H4 = (C=C) + 4(C-H) = 614 + 4(C-H) = 2658.1

(C-H) = (2658.1 - 614)/4 = 511.05 KJ/mol

c) C2H2 + 3H2 ------> 2CH4 ΔH = -376.74 KJ/mol

Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol

Bond energy of reactants = Bond energy of C2H2 + (3 × Bond energy of H2)

Bond energy of C2H = (triple bond Carbon to Carbon) + 2(C-H) = 839 + 2(C-H)

(3 × Bond energy of H2) = 3 × 432 = 1296 KJ/mol

-376.74 = 3320 - (Bond energy of C2H2 + 1296)

Bond energy of C2H2 = 3320+376.74-1296 = 2400.74 KJ/mol

Bond energy of C2H2 = (triple bond Carbon to Carbon) + 2(C-H) = 839 + 2(C-H) = 2400.74

(C-H) = (2400.74 - 839)/2 = 780.87 KJ/mol

QED!!!

6 0

3 years ago