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2 years ago

The density (in g/cm3) of a gold nugget that has a volume of 1.68 cm3 and a mass of 32.4 g is __________.

2 answers:

6 0

The density of a substance is extensive even though its unit components, mass and volume, are intensive properties. At a certain temperature and pressure, the density of a substance is constant. It is the ratio of the substance's mass to volume.

Density = Mass/Volume
Density = 32.4 g/1.68 cm³
Density = 19.29 g/cm³

Drupady [299]2 years ago

5 0

Answer:

Density of the gold nugget, $\rho=19.28\ g/cm^3$

Explanation:

It is given that,

Mass of the gold nugget, m = 32.4 g

Volume of the gold nugget $V=1.68\ cm^3$

We need to find the density of the gold nugget. The total mass divided by total volume of an substance is called its density. Its formula is given by :

$\rho=\dfrac{m}{V}$

$\rho=\dfrac{32.4\ g}{1.68\ cm^3}$

$\rho=19.28\ g/cm^3$

So, the density of the metal ball is $19.28\ g/cm^3$ . Hence, this is the required solution.

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Hydrogen peroxide, H2O2, is common in many households and is used as a bleaching agent. It usually comes in a dark, opaque bottl

sergij07 [2.7K]

Answer:

Hydrogen peroxide should be stored in

1) a cool environment

2) with amber bottles away from sunlight

3) with little drops of sodium phosphate

Explanation:

It has been confirmed that heat and light aids in the decomposition of hydrogen peroxide according to the equation; 2H2O2→2 H2O + O2.

This means that hydrogen peroxide must be stored in a cool place. This will reduce its rate of decomposition. Secondly, it should be stored in amber bottles away from light since light also aids in its decomposition.

Thirdly, drops of sodium phosphate may be added to prevent its catalytic decomposition during storage.

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2 years ago

A 0.500 g sample of tin (Sn) is reacted with oxygen to give 0.534 g of product. What is the empirical formula of the oxide?

REY [17]

Answer:

$Sn_2O$

Explanation:

Hello,

In this case, given that the mass of the product is 0.534 g, we can infer that the percent composition of tin is:

$\%Sn=\frac{0.500g}{0.534g}*100\%\\ \\\%Sn=93.6\%$

Therefore, the percent composition of oxygen is 6.4% for a 100% in total. Thus, with such percents we compute the moles of each element in the oxide:

$n_{Sn}=93.6gSn*\frac{1molSn}{118.8gSn} =0.788molSn\\\\n_O=6.4gO*\frac{1molO}{16gO}=0.4molO$

In such a way, for finding the smallest whole number we divide the moles of both tin and oxygen by the moles of oxygen as the smallest moles:

$Sn:\frac{0.788}{0.4}=2\\ \\O:\frac{0.4}{0.4}=1$

Therefore, the empirical formula is:

$Sn_2O$

Best regards.

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2 years ago

Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the exces

Shalnov [3]

Answer: The concentration of excess $[OH^-]$ in solution is 0.017 M.

Explanation:

1. $Molarity=\frac{moles}{\text {Volume in L}}$

moles of $HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles$

1 mole of $HCl$ give = 1 mole of $H^+$

Thus 0.019 moles of $HCl$ give = 0.019 mole of $H^+$

2. moles of $Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles$

According to stoichiometry:

1 mole of $Ba(OH)_2$ gives = 2 moles of $OH^-$

Thus 0.012 moles of $Ba(OH)_2$ give = $2 \times 0.012=0.024$ moles of $OH^-$

$H^++OH^-\rightarrow H_2O$

As 1 mole of $H^+$ neutralize 1 mole of $OH^-$

0.019 mole of $H^+$ will neutralize 0.019 mole of $OH^-$

Thus (0.024-0.019)= 0.005 moles of $OH^-$ will be left.

$[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M$

Thus molarity of $[OH^-]$ in solution is 0.017 M.

4 0

2 years ago

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When balanced, which equation would have the coefficient 3 in front of any of the reactants? Zn + HCl ZnCl2 + H2 H2SO4 + B(OH)3

Gwar [14]

The correct answer is B. H2SO4 + B(OH)3 B2(SO4)3 + H2O

Hope this helps!

6 0

2 years ago

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Find the volume of a gas at STP, if its volume is 80.0 mL at 109 kPa and -12.5°C.

exis [7]

Answer:

= 913.84 mL

Explanation:

Using the combined gas laws

P1V1/T1 = P2V2/T2

At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.

V1 = 80.0 mL

P1 = 109 kPa

T1 = -12.5 + 273 = 260.5 K

P2 = 10 kPa

V2 = ?

T2 = 273 K

Therefore;

V2 = P1V1T2/P2T1

= (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)

<u>= 913.84 mL</u>

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2 years ago

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