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jonny [76]
2 years ago
15

Density = _____ (A)weight/length (B)mass/weight (C)mass/volume (D)volume/weight

Chemistry
2 answers:
8_murik_8 [283]2 years ago
7 0
Density= Mass/Volume I am positive I just had an assignment on this
algol132 years ago
5 0
C mass and volume i remember by d=m/v
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The empirical formula for carbohydrates is CH2O. The molecular weight for a particular carbohydrate is 90.09 grams. What is the
baherus [9]
We find the weight of the empirical formula:
12.0107 + 2 x 1.00794 + 15.9994
= 30.03

Now, we divide the molecular weight by the weight of the empirical formula to find the number of times the empirical formula repeats:
90.09 / 30.03
= 3

The formula is 3(CH₂O)
C₃H₆O₃
6 0
3 years ago
What is the formula weight (amu) of the molecule H2O? Use atomic masses of H and O as 1.008 amu and 16.00 amu respectively. Repo
vaieri [72.5K]

Answer:

Formula weight of H₂O molecule is  18.02 amu.

Explanation:

Given data:

Formula weight of H₂O = ?

Atomic mass of H = 1.008 amu

Atomic mass of O = 16.00 amu

Solution:

Formula weight:

"It is the sum of all the atomic weight of atoms present in given formula"

Formula weight of H₂O = 2×1.008 amu + 1×16.00 amu

Formula weight of H₂O = 18.02 amu

Thus, formula weight of H₂O molecule is  18.02 amu.

6 0
3 years ago
Help on my homework
Illusion [34]

Bronchi

Bronchioles

Alveoli

3 0
3 years ago
When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be
Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH  = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL  +  50.mL  = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH_{neutralization}

ΔH_{neutralization} = -q_soln / mole of water produced

so we substitute

ΔH_{neutralization} = -( 4076.68 J ) / 0.0500 mol  

ΔH_{neutralization} = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

6 0
2 years ago
Please help!!
prisoha [69]
A word equation is a written description of a chemical reaction.
All word equations start with the reactants.
Then, what comes next is the word "react to form."
Finally, the products of the equation are mentioned.
An example is,
Zinc and Hypochondriac acid react to form hydrogen gas and zinc chloride.
3 0
3 years ago
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