47526 to the nearest 10

What is the volume of 3.5 moles of oxygen gas (O2) at standard temperature and pressure (STP)?

baherus [9]

1 mole ----------- 22.4 L ( at STP )
3.5 mols --------- ?

V = 3.5 x 22.4 / 1

V = 78.4 L

hope this helps!

4 0

3 years ago

How many are molecules ( or formula) in each sample?

andre [41]

Answer:

$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

Explanation:

Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

8 0

3 years ago

How do balanced chemical equations show the conservation of mass

antiseptic1488 [7]

Answer: Matter cannot be created or destroyed

Explanation: Balanced equations are set equations we cannot change one element or compound in the equation without changing the entire equation. So balanced equation show the conservation of mass because while other substances may be formed from the synthesis or decomposition of compounds new elements are never introduced and are not created out of thin air :)

4 0

3 years ago

Read 2 more answers

How many of grams of CuSO4 are in 475ml of a 2.0M aqueous solution

Anettt [7]

Answer:

151.63 g

Explanation:

We first get the number of moles;

Moles = Molarity × volume

= 2.0 × 0.475

= 0.95 moles

1 mole of CuSO4 = 159.609 g/mol

Therefore;

Mass = moles × molar mas

= 0.95 moles × 159.609 g/mol

= 151.63 g

8 0

4 years ago

Read 2 more answers

What type of solution has a pH of 7?

motikmotik

If pH+ is 7
It is Neutral solution

5 0

4 years ago

47526 to the nearest 10​

47526 to the nearest 10