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vesna_86 [32]
3 years ago
9

What would the word equation be?

Chemistry
1 answer:
valentinak56 [21]3 years ago
5 0
The problem=the equation
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What are cells made of?
skad [1K]

Answer: All cells are made from the same major classes of organic molecules: nucleic acids, proteins, carbohydrates, and lipids.

Explanation: please mark as brainliest

8 0
3 years ago
Linolenic acid (C18H30O2 - M.W. = 278.42 g/mol) reacts with hydrogen gas according to the equation: C18H30O2 + 3H2 (g) → C18H36O
Ludmilka [50]

Answer:

2.53 L is the volume of H₂ needed

Explanation:

The reaction is: C₁₈H₃₀O₂ + 3H₂ → C₁₈H₃₆O₂

By the way we can say, that 1 mol of linolenic acid reacts with 3 moles of oxygen in order to produce, 1 mol of stearic acid.

By stoichiometry, ratio is 1:3

Let's convert the mass of the linolenic acid to moles:

10.5 g . 1 mol / 278.42 g  = 0.0377 moles

We apply a rule of three:

1 mol of linolenic acid needs 3 moles of H₂ to react

Then, 0.0377 moles will react with (0.0377 . 3 )/1 = 0.113 moles of hydrogen

We apply the Ideal Gases Law to find out the volume (condition of measure are STP) → P . V = n . R . T → V = ( n . R .T ) / P

V = (0.113 mol . 0.082 L.atm/mol.K . 273.15K) 1 atm = 2.53 L

4 0
3 years ago
The half-life for the second-order decomposition of HI is 15.4 s when the initial concentration of HI is 0.67 M. What is the rat
larisa [96]

Answer:

The correct answer is option C.

Explanation:

Half life for second order kinetics is given by:

t_{\frac{1}{2}=\frac{1}{k\times A_0}

t_{\frac{1}{2} = half life = 15.4 s

k = rate constant =?

A_0 = initial concentration = 0.67 M

15.4 s=\frac{1}{k\times 0.67 M}

k=\frac{1}{15.4 s\times 0.67 M}=0.09692 M^{-1} s^{-1}=9.69\times 10^{-2}M^{-1} s^{-1}\approx 9.7\times 10^{-2} M^{-1} s^{-1}

9.7\times 10^{-2} M^{-1} s^{-1}  is the rate constant for this reaction.

4 0
3 years ago
1. Number of planets in our solar system.<br> A. 6<br> B. 7<br> C. 8<br> D. 9
satela [25.4K]

Answer:

8

Explanation:

Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune all 8 planets

7 0
3 years ago
Read 2 more answers
the coal sample being used in this experiment is assumed to have 3% sulfur in its composition. what percent of sulfate is presen
mote1985 [20]

Answer: The percent of sulfate is present in the coal samples is 0.999%.

Explanation:

Given: Percentage of sulfur present = 3%

In SO^{2-}_{4}, the percentage by weight of Sulfur is calculated as follows.

Percentage = \frac{32 g}{32 + 64} \times 100\\= 33.33 percent

This means that the amount of only sulfur present in SO^{2-}_{4} in the sample is as follows.

\frac{3}{100} \times 33.33\\= 0.999 percent

Thus, we can conclude that percent of sulfate is present in the coal samples is 0.999%.

7 0
3 years ago
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