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Goshia [24]
3 years ago
13

Help please, thank you:)

Chemistry
1 answer:
IgorC [24]3 years ago
6 0

Answer:

1.67 × 10⁻¹⁰m is the answer.

Explanation:

Given,

4 × 10⁻⁷m thick

2400 atoms thick

therefore 1 atom,

Diameter = 4 × 10⁻⁷/2400

d = 4000 × 10⁻¹⁰/2400

d = 1.67 × 10⁻¹⁰m

∴ diameter of 1 gold atom (3 significant figures) = 1.67 × 10⁻¹⁰m

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Answer:

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One of relatively few reactions that takes place directly between two solids at room temperature is Ba(OH)2 · 8H2O + ammonium th
Hatshy [7]

Answer:

\boxed{\text{3.1 g}}

Explanation:

We will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:           315.46               76.12

           Ba(OH)₂·8H₂O + 2NH₄SCN ⟶ Ba(SCN)₂ + 2NH₃ + 10H₂O

m/g:             6.5

1. Moles of Ba(OH)₂·8H₂O

\text{Moles of Ba(OH)$_{2}\cdot $8H$_{2}$O}\\= \text{ 6.5 g Ba(OH)$_{2}\cdot $8H$_{2}$O} \times \dfrac{\text{1 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}}{\text{ 315.46 g Ba(OH)$_{2}\cdot $8H$_{2}$O}}\\= \text{0.0206 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}

2. Moles of NH₄SCN

The molar ratio is 2 mol NH₄SCN:1 mol Ba(OH)₂·8H₂O

\text{Moles of NH$_{4}$SCN} =\text{0.0206 mol Ba(OH)$_{2}\cdot $8H$_{2}$O} \times \dfrac{\text{2 mol NH$_{4}$SCN}}{\text{1 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}}\\= \text{0.0412 mol NH$_{4}$SCN}

3. Mass of NH₄SCN

\text{Mass of NH$_{4}$SCN} = \text{0.0412 mol NH$_{4}$SCN} \times \dfrac{\text{76.12 g NH$_{4}$SCN}}{\text{1 mol NH$_{4}$SCN}} =\\\textbf{3.1 g NH$_{4}$SCN}\\\\\text{You must use }\boxed{\textbf{3.1 g}}\text{ NH$_{4}$SCN}

5 0
4 years ago
Why is chlorine a gas
natulia [17]
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Chlorine<span> is a </span>gas at room temperature.<span>The melting/boiling point of a substance determines what state of matter it takes at a certain temperature.</span> 
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3 years ago
C2H6O(g) + 3O2(g) 2CO2(g) + 3H2O(g) Inside the piston of an automobile engine, 0.461 g of ethanol (C2H6O) gas reacts with 0.640
LuckyWell [14K]

Answer:

Oxygen is the limiting reactant.

Explanation:

Hello,

In this case, given the reaction:

C_2H_6O(g) + 3O_2(g)\rightarrow 2CO_2(g) + 3H_2O(g)

Hence, given the masses of both ethanol and oxygen, we are able to compute the available moles ethanol by:

n_{C_2H_6O}^{available}=0.461g*\frac{1mol}{46g}=0.01mol C_2H_6O

Next, we compute the moles of ethanol that react with the 0.640 grams of oxygen considering their 1:3 molar ratio in the chemical reaction:

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In such a way, since there are 0.01 available moles of ethanol but just 0.0067 moles are reacting, we evidence ethanol is in excess, therefore the oxygen is the limiting reactant.

Best regards.

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3 years ago
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