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GREYUIT [131]
3 years ago
8

How heavy is one atom of hydrogen

Chemistry
1 answer:
Harman [31]3 years ago
7 0

Hello!

the atomic mass of one atom of hydrogen is equal to 1.00784 in atomic mass units

one atomic mass unit is equal to 3.6609e-27 pounds

which means one atom of hydrogen is equal to 3.9478732e-27 pounds

I hope this helps, and have a nice day!

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The main two characteristics that describe any igneous rock are
amm1812
All igneous rocks are defined by texture and gas content.  Hope this helped, my man.
7 0
3 years ago
What is the importance of "Sodium" for human health?​
Aneli [31]

Answer:

Explanation:

Sodium is both an electrolyte and mineral. It helps keep the water (the amount of fluid inside and outside the body's cells) and electrolyte balance of the body. Sodium is also important in how nerves and muscles work. Most of the sodium in the body (about 85%) is found in blood and lymph fluid.

6 0
3 years ago
Lead (II) carbonate decomposes to give lead (II) oxide and carbon dioxide: PbCO 3 (s) PbO (s) CO 2 (g) ________ grams of lead (I
blsea [12.9K]

Answer:

We will have 7.30 grams lead(II) oxide

Explanation:

Step 1: Data given

Mass of lead (II)carbonate = 8.75 grams

Molar mass PbCO3 = 267.21 g/mol

Step 2: The balanced equation

PbCO3 (s) ⇆ PbO(s) + CO2(g)

Step 3: Calculate moles PbCO3

Moles PbCO3 = mass / molar mass

Moles PbCO3 = 8.75 grams / 267.21 g/mol

Moles PbCO3 = 0.0327 moles

Step 4: Calculate moles PbO

For 1 mol PbCO3 we'll have 1 mol PbO and 1 mol CO2

For 0.0327 moles PbCO3 we'll have 0.0327 moles PbO

Step 5: Calculate mass PbO

Mass PbO = moles PbO * molar mass PbO

Mass PbO = 0.0327 moles * 223.2 g/mol

Mass PbO = 7.30 grams

We will have 7.30 grams lead(II) oxide

7 0
4 years ago
What percent of I-125 has decayed if there are 37.5g of the original sample left?
garri49 [273]

Answer:

[(x-37.5)/x]*100%

62.5%  (assuming the original sample weighs 100.0g)

Explanation:

Let's say that the original sample is x

Mass of I-125 which has decayed: x-37.5

Percentage of decayed mass: [(x-37.5)/x]*100%

Please recheck, for this may not be the correct answer

6 0
3 years ago
Is the reaction to produce zinc from zinc sulfide spontaneous under standard conditions? Coupledreaction: ZnS(s) + H2 (g)  Zn(s
finlep [7]

Answer:

The reaction to produce zinc from zinc sulfide is not spontaneous under standard conditions.

Explanation:

Hess's law states: "When one reaction can be expressed as the algebraic sum of others, its heat of reaction is equal to the same algebraic sum of the partial heats of the partial reactions". So Hess's Law is an indirect method of calculating the heat of reaction or enthalpy of reaction when the chemical reaction occurs in one or more than one stage.

Taking into account that ΔG is a state function, which only depends on the initial and final states, its variation in a reaction is calculated by adding the free energies of the reactants and products involved in it when both are in the normal state. That is, at the pressure of 1natm if it is gases or at the concentration of 1 mol / L for substances in liquid solution.

The sum of the fitted equations should give the problem equation. So, if in a "data" reaction a substance is as a reactant and in the reaction that you must obtain is as a product, you must turn the "data" reaction and ΔG will change its sign. In this case, this happen with the reaction 1 (Rxn1):

Rxn1: ZnS (s) → Zn (s) + S (s)          ΔG1°= 201.3 kJ

Rxn2: S (s) + H₂ (g) → H₂S (g)        ΔG2°= -33.4kJ

Adding both reactions (taking into account that certain substances appear sometimes as a reagent and others as a product, so they are totally eliminated if they appear in the same quantities) you get:

ZnS (s) + H₂ (g) → Zn (s) + H₂S (g)

Adding algebraically ΔG1° and ΔG2° you get:

ΔG°= ΔG1° + ΔG2°

ΔG°= 201.3 kJ - 33.4 kJ

ΔG°= 167.9 kJ

If a chemical reaction proceeds with a ΔG <0 the process is  spontaneous. If, on the other hand, ΔG> 0, the reaction is not spontaneous.

Since ΔG>0, <u><em>the reaction to produce zinc from zinc sulfide is not spontaneous under standard conditions.</em></u>

6 0
3 years ago
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