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3 years ago

15

the density of gold is 19.3 g/cm the density of iron pyrite is 5.0 g/cm . if a nugget of iron pyrite and a nugget of gold each h

ave a mass of 50 g, what can you conclude about the volume of each nugget

1 answer:

UkoKoshka [18]3 years ago

4 0

Iron Pyrite Volume=10mL

Gold Volume=2.59mL

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How to change 5 % W/V of NaCl to ppm , M ? molar mass = 58.5<br>please clear explain

11Alexandr11 [23.1K]

Answer:

50000ppm and 0.855M.

Explanation:

ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters

A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.

To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:

<em>mg NaCl:</em>

5g * (1000mg / 1g) = 5000mg

<em>L Solution:</em>

100mL * (1L / 1000mL) = 0.100L

ppm:

5000mg / 0.100L = 50000ppm

To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:

5g * (1mol / 58.5g) = 0.0855moles NaCl

Molarity:

0.0855mol NaCl / 0.100L = 0.855M

7 0

3 years ago

Plants that have nigrogen fixing bacteria in their roots are called

iragen [17]

Plants that have nigrogen fixing bacteria in their roots are called

legumes.

8 0

3 years ago

The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?

Sloan [31]

Answer: The concentration of hydrogen ions for this solution is $1.99 \times 10^{-3}$ .

Explanation:

Given: pOH = 11.30

The relation between pH and pOH is as follows.

pH + pOH = 14

pH + 11.30 = 14

pH = 14 - 11.30

= 2.7

Also, pH is the negative logarithm of concentration of hydrogen ions.

$pH = - log [H^{+}]$

Substitute the values into above formula as follows.

$pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}$

Thus, we can conclude that the concentration of hydrogen ions for this solution is $1.99 \times 10^{-3}$ .

4 0

3 years ago

A student collected a 47.5 mL sample of gas in the lab at 0.8 atm pressure and 29.00C. What volume would this gas sample occupy

Sav [38]

Answer:

the volume would be 69.034mL

Explanation:

8 0

2 years ago

Can someone help me? It needs to have a diagram that has arrows.

daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]$

Putting the values we get :

$\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]$

$-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]$

$H_f_{C_4H_{10}=-125kJ/mol$

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0

3 years ago