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1 year ago

How many moles of copper are in 6,000,000 atoms of copper?

Chemistry

1 answer:

crimeas [40]1 year ago

7 0

Answer: There will be 9.9632 × 10⁻¹⁸ moles of Copper in 6,000,000 atoms of Copper.

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PLEASE HELP!!! what occurs along a convergent plate boundary

forsale [732]

your answer is c. your welcome

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2 years ago

Read 2 more answers

Methanol can be produced by the following reaction: CO(g) 2 H2(g) CH3OH(g). How is the rate of disappearance of hydrogen gas rel

belka [17]

Answer:

$r_{H_2}=-2r_{CH_3OH}$

Explanation:

Hello!

In this case, for the reaction:

$CO(g)+ 2 H_2(g) \rightarrow CH_3OH(g)$

In such a way, via the rate proportions, that is written considering the stoichiometric coefficients, we obtain:

$\frac{1}{-1} r_{CO}=\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}$

Whereas the reactants, CO and H2 have negative stoichiometric coefficients; therefore the rate of disappearance of hydrogen gas is related to the rate of appearance of methanol as shown below:

$\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}\\\\r_{H_2}=\frac{-2}{1} r_{CH_3OH}\\\\r_{H_2}=-2r_{CH_3OH}$

Which means that the rate of disappearance of hydrogen gas is negative and the rate of appearance of methanol is positive.

Regards!

7 0

2 years ago

Calculate the volume in dm3 occupied by 26.4 g of carbon dioxide

Mama L [17]

Density of carbon dioxide=1564kg/m^3
Mass=26.4g=0.0264kg

$\\ \sf\longmapsto Density=\dfrac{Mass}{Volume}$

$\\ \sf\longmapsto Volume=\dfrac{Mass}{Density}$

$\\ \sf\longmapsto Volume=\dfrac{0.0264}{1564}$

$\\ \sf\longmapsto Volume=1.687m^3=1.69m^3$

1m^3=1000dm^3

$\\ \sf\longmapsto Volume=1.69(1000)=1690dm^3$

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2 years ago

A 10.0-mL solution of 0.30 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the

yulyashka [42]

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2 years ago

Given that the density of water is 0.975 g/ml and that 171 g of sucrose (molar mass: 342.30 g/mol) is dissolved in 512.85 ml of

VMariaS [17]

Molality is defined as the number of moles of solute in 1 kg of solvent .
the mass of water - 512.85 ml x 0.975 g/ml = 500 g
number of moles of solute in 500 g of water - 171 g / 342.30 g/mol = 0.500 mol
the number of sucrose moles in 500 g of water - 0.500 mol
therefore number of sucrose moles in 1 kg water - 0.500 mol / 0.500 kg = 1.00
molality of solution - 1.00 M

6 0

2 years ago