All categories

3 years ago

Work done after rutherford's gold foil experiment demonstrated that the nucleus contained ________.

Chemistry

1 answer:

shutvik [7]3 years ago

5 0

Protons and neutrons

hope this helps

You might be interested in

78.5 mol of P4O10 contains how many moles of P?

MArishka [77]

4 x 36 moles. of Phosporus and 10 x 36 of Oxygen. I hope this helps. (:

7 0

3 years ago

Which statement is true about the exothermic reaction?

Stells [14]

Answer:

Explanation:

The enthalpy of the reactants is greater than that of the products.

7 0

2 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

Answer both 8 and 9 will give u brain list with explanation as well so don’t just answer

goldfiish [28.3K]

Answer:

8. the answer is B.

9. the answer is A.

Explanation:

Hello!

8. In this case, by bearing to mind that the limiting reactant is always completely consumed and the excess one remain as a leftover at the end of the reaction, we can also infer that as all the limiting reactant is consumed, it must determine the maximum amount of product as the excess reactant will hypothetically produce a greater mass than expected; thus, the answer to this question is B.

9. In this case, since the mole ratio of oxygen to water is 1:2, the following proportional factor is used to calculate the produced mass of water:

$3molO_2*\frac{2molH_2O}{1molO_2}=6molH_2O$

Thus, the answer is this case is A.

Best regards!

6 0

2 years ago

aluminum bromide reacts with chlorine gas to produce aluminum chloride and bromide gas. if we have 9 moles of chlorine gas. how

Luda [366]

Moles of Bromine produced = 9 moles

<h3>Further explanation</h3>

Given

9 moles of Chlorine gas

Word equation

Required

Moles of Chlorine produced

Solution

We change the word equation into a chemical equation (with a formula)

Aluminum bromide reacts with chlorine gas to produce Aluminum chloride and bromide gas

2AlBr₃+3Cl₂⇒2AlCl₃+3Br₂

moles Cl₂ = 9

Maybe you mean, how many moles of Bromine can we produce?

From equation, mol ratio Cl₂ : Br₂ = 3 : 3, so mol Br₂=mol Cl₂=9 moles

7 0

3 years ago