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2 years ago

How would one convert 15 seconds to hours?

Chemistry

2 answers:

laiz [17]2 years ago

8 0

Explanation:

a ...15 seconds×(3600 seconds hour)

frozen [14]2 years ago

3 0

15 seconds x 1 hour/3600 seconds

You might be interested in

Are diamonds flammable?

Debora [2.8K]

A match-head is flammable, but you have to get it up to its ignition temperature first.

A diamond is flammable, but its ignition temperature is astronomical and
totally out of sight.

4 0

3 years ago

Read 2 more answers

According to the Law of Conservation of Mass, how much water is produced if 22.5 g of

nadezda [96]

Answer: The amount of water produced is 9.3 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

mass of reactants = mass of methane + mass of oxygen = 22.5 g + 35.7 g = 58.2 g

mass of products = mass of carbon dioxide + mass of water = 48.9 g + mass of water

48.9 g + mass of water = 58.2 g

mass of water = 9.3 g

4 0

3 years ago

What is the mass of 7.50 moles of magnesium chloride, mgcl2?

Natali5045456 [20]

MgCl2 = 1Mg + 2Cl = 1(24.3) + 2(35.45) = 95.2g/1mole
7.50moles MgCl2 x 95.2g MgCl2 = 714g MgCl2

8 0

3 years ago

Dissolve 30 g of sodium sulphate into 300 mL of water

Aneli [31]

Answer:

number of moles = 0.21120811

Explanation:

To find the number of moles, given the mass of the solute, we use the formula:

$\mathrm{n = \dfrac{ m }{ M } }$

$\mathrm{n = number\:of\:moles\:(mol)}$

$\mathrm{m = mass\:of\:solute\:(g)}$

$\mathrm{M = molar\:mass\:of\:solute\:( \dfrac{ g }{ mol } )}$

Label the variables with the numbers in the problem:

$\mathrm{n =\:?}$

$\mathrm{m =30\:g }$

$\mathrm{M =\:?\:Calculate\:the\:molar\:mass }$

The first thing we have to do is find the molar mass of sodium sulfate, in order for us to use the formula for finding the number of moles:

Formula for finding the molar mass of sodium sulfate:

$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

For the variables and what they mean are below for finding the molar mass of sodium sulfate:

$\mathrm{M =molar\:mass }$

$\mathrm{m =moles=2\:moles\:for\:Na\:,1\:mole\:for\:S,\:and\:4\:moles\:for\:O}$

$\mathrm{Na =sodium=22.99\:g }$

$\mathrm{S =sulfur=32.06\:g }$

$\mathrm{O =oxygen=16.00\:g }$

Plug the numbers into the formula, to find the molar mass of sodium sulfate:

$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

$M = 2 \left( 22.99 \right) +1 \left( 32.06 \right) +4 \left( 16.00 \right)$

$\mathrm{Multiply\:2\:by\:22.99\:to\:get\:45.98\:and\:1\:by\:32.06\:to\:get\:32.06}$

$\mathrm{M = 45.98+32.06+4\:(16)}$

$\mathrm{Multiply\:4\:by\:16\:to\:get\:64}$

$\mathrm{M = 45.98+32.06+64}$

$\mathrm{Add\:45.98\:and\:32.06\:to\:get\:78.04}$

$\mathrm{M = 78.04+64}$

$\mathrm{Add\:78.04\:and\:64\:to\:get\:142.04}$

$\mathrm{M = 142.04}$

Now that we have found the molar mass, we can calculate the number of moles in the solution of sodium sulfate with the formula:

$\mathrm{n = \dfrac{ m }{ M } }$

$\mathrm{n =\:?}$

$\mathrm{m =30\:g }$

$\mathrm{M = 142.04\:g/mol}$

$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

$\mathrm{n = \dfrac{ 30 }{ 142.04 }}$

$\mathrm{Divide\:142.04\:by\:30\:to\:get\:0.21120811}$

$\mathrm{n = 0.21120811}$

0.21120811 rounded gives you 0.2112

or if you did the problem without decimals

30 grams of sodium sulfate divided by its molecular weight – which we found to be 142 – gives us a value of 0.2113 moles.

3 0

2 years ago

Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed leave it blank. If no reaction occurs leav

Andrei [34K]

Answer:

ZnS(s) ⇄ S²⁻(aq) + Zn²⁺(aq)

Explanation:

First, we will write the molecular equation, since it is easier to balance.

2 HBr(aq) + ZnS(s) ⇄ H₂S(aq) + ZnBr₂(aq)

In the full ionic equation we include all ions and molecular species.

2 H⁺(aq) + 2 Br⁻(aq) + ZnS(s) ⇄ 2 H⁺(aq) + S²⁻(aq) + Zn²⁺(aq) + 2 Br⁻(aq)

In the net ionic equation we include only the ions that participate in the reaction and the molecular species.

ZnS(s) ⇄ S²⁻(aq) + Zn²⁺(aq)

6 0

3 years ago