How many moles of MgSiO3 are in 237g of the compound?

A student creates a model that displays a cross-section of a metal wire wrapped in plastic insulation. Which statement best desc

artcher [175]

Cross-section of a metal wire wrapped in plastic insulation the metal wire is an element.

<h3>What is an element?</h3>

An element is a substance or material that cannot be affected by an external substance or cannot be broken down by any reaction.

Metal wire is serving as an element, it should not be easily broken or penetrated as an insulator.

Therefore, Cross-section of a metal wire wrapped in plastic insulation the metal wire is an element.

Learn more on element here,

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7 0

3 years ago

The conversion of cyclopropane to propene occurs with a first-order rate constant of . How long will it take for the concentrati

Gennadij [26K]

This is an incomplete question, here is a complete question.

The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

Answer : The time taken will be, 17.0 hr

Explanation :

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $2.42\times 10^{-2}\text{ hr}^{-1}$

t = time passed by the sample = ?

a = initial concentration of the reactant = 0.080 M

a - x = concentration left = 0.053 M

Now put all the given values in above equation, we get

$t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}$

$t=17.0\text{ hr}$

Therefore, the time taken will be, 17.0 hr

5 0

3 years ago

2) what is the mass of 6.02 x 1023 atoms of arsenic?

zzz [600]

Data:
Arsenic Molar Mass = 74,9216 ≈ 75 u (<span>atomic mass unit)</span>

Solving:

1 mole of arsenic → 75g ------------ 6,02*10²³ molecules
..................................X -------------- 1 molecule

6,02*10²³X = 75
$x = \frac{75}{6,02*10^{23}}$
$\boxed{x \approx 1,24*10^{24}grams}$

6 0

3 years ago

How many grams of water (H2O) will be produced from 96.0 grams of oxygen (O2)?

jekas [21]

Answer & Explenation:

2H2 + O2 -> 2H20 is the balanced equation for the reaction of Hydrogen with Oxygen to form water so

If you have 32g of O2 this is a simple Dimensional analysis problem

32g O2 x 36.03056g H20/31.9988g O2 this way the O2 cancels out and you are left with just the H2O so your raw answer would be 36.0319112, then if your instructor requires a significant figure answer that would be to 2 significant figures the information you were given 32g O2, so as above 36g or Water are produced. Just a different way to view and solve the problem with the balanced equation so you can see the way everything relates to everything else. the molar masses of O2 and H2O are simply found my adding up 2 Oxygens 15.9994g x2 = 31.9988g and H2O = 2(1.00794) + 15.9994 = 18.01258 but you then have to multiply that by 2 because the reaction states you get 2 mols in the reaction so that is where I came up with the 36.03056g for the solution. Hope this helps. Dimensional Analysis is and can be one of the best ways to solve these problems, because not always are you going to be dealing with 1:2 ratios.

4 0

3 years ago

Considering the volumes and the concentrations used in mixture 1, what percentage of the moles of H2O2 present have been consume

Kisachek [45]

Answer:

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

Explanation:

the total volume of the mixture is equal to:

Vmix = 75 + 30 + 25 + 5 + 5 + 10 = 150 mL

the moles of each species in the mix equals:

Na2S2O3 = 0.05 * 0.005 = 0.00025 moles

KI = 0.05 * 0.025 = 0.00125 moles

H2O2 = 1.02 * 0.01 = 0.0102 moles

the following equation shows the reaction between I2 and S2O32:

I2 + S2O32 = 2I- + S4O62-

The same way:

2I- + 2H+ + H2O2 = I2 + 2H2O

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

3 0

3 years ago