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Ulleksa [173]
3 years ago
15

A mountain has a height of 2.74 miles. How high is the mountain in meters? Use the fact that 1 mi=1.609 km .

Chemistry
1 answer:
serious [3.7K]3 years ago
6 0

kkknskndcnndcks

a

Answer:

Explanation:

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Find the volume of hydrogen gas formed when 1.5g of aluminum reacts with aq NaOH at 27 degrees Celcius.
Rudiy27

Answer:

how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?

(A) n=m/M,

n(Al)=5.4/27=0.2 moles

n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles

Number of oxygen atoms= n(O2)*Avogadro's number

=0.15*6.02*10^23=9.03*10^22 oxgyen atoms

(B)

n=m/M

n(Al)=0.6/27=0.02222 moles

n(O2)=n(Al)*3/4=0.016666 moles

m=n*M

m(O2)=0.0166666*32=0.53333 grams

8 0
2 years ago
8
love history [14]

Answer : The molar concentration of sucrose in the tea is, 0.0549 M

Explanation : Given,

Mass of sucrose = 3.765 g

Volume of solution = 0.200 L

Molar mass of sucrose  = 342.3 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of sucrose}}{\text{Molar mass of sucrose}\times \text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{3.765g}{342.3g/mole\times 0.200L}=0.0549mole/L=0.0549M

Therefore, the molar concentration of sucrose in the tea is, 0.0549 M

7 0
3 years ago
Which of the following reactions could be used to extract the lead from the lead nitrate?
alexgriva [62]

Answer:

Option b. Decomposition

Followed by a reduction process using charcoal

Explanation:

Lead can be obtained from lead nitrate by thermal decomposition of lead nitrate as shown below:

2Pb(NO3)2 —> 2PbO + 4NO2 + O2

The PbO obtained is reduced by charcoal(C) to obtain the metallic Pb as shown below:

2PbO + C —> Pb + CO2

4 0
3 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
M=10,V=2 What is the density
Sever21 [200]

Answer:

Density (ρ) = 5 kilogram/cubic meter

Explanation:

Steps:

ρ =  

m

V

=  

10 kilogram

2 cubic meter

=  5 kilogram/cubic meter

4 0
3 years ago
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