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1 year ago

Calculate the mass (in grams) of 250mL of ether at 25 oC. The density of

Chemistry

1 answer:

leonid [27]1 year ago

6 0

Volume=250mL
Density=0.71g/ml

$\boxed{\sf Density=\dfrac{Mass}{Volume}}$

$\\ \sf{:}\implies Mass=Density(Volume)$

$\\ \sf{:}\implies Mass=0.71(250)$

$\\ \sf{:}\implies Mass=177.5g$

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A gold wire has a diameter of 1.00 mm. What length of this wire contains exactly 1.00 mol of gold? (density of Au = 17.0 g/cm3)

allochka39001 [22]

Answer:

The answer to your question is 7160 cm

Explanation:

Data

diameter = 1 mm

length = ?

amount of gold = 1 mol

density = 17 g/cm³

Process

1.- Get the atomic mass of gold

Atomic mass = 197 g

then, 197g ------------ 1 mol

2.- Calculate the volume of this wire

density = mass/volume

volume = mass/density

volume = 197/17

volume = 5.7 cm³

3.- Calculate the length of the wire

Volume = πr²h

solve for h

h = volume /πr²

radius = 0.05 cm

substitution

h = 5.7/(3.14 x 0.05²)

h = 5.7 / 0.0025

h = 7159.2 cm ≈ 7160 cm

8 0

2 years ago

Read 2 more answers

A solution contains 0.0150 M Pb2+(aq) and 0.0150 M Sr2+(aq) . If you add SO2−4(aq) , what will be the concentration of Pb2+(aq)

AnnyKZ [126]

Answer:

$\large \boxed{1.10 \times 10^{-3}\text{ mol/L}}$

Explanation:

1. Concentration of SO₄²⁻

SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷

0.0150 x

$K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}$

2. Concentration of Pb²⁺

PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸

x 2.293 × 10⁻⁵

$K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}$

4 0

2 years ago

Acetaldehyde decomposes at 750 K: CH3CHO → CO + CH4. The reaction is first order in acetaldehyde and the half-life of the reacti

Degger [83]

Answer:

k = 1.3 x 10⁻³ s⁻¹

Explanation:

For a first order reaction the integrated rate law is

Ln [A]t/[A]₀ = - kt

where [A] are the concentrations of acetaldehyde in this case, t is the time and k is the rate constant.

We are given the half life for the concentration of acetaldehyde to fall to one half its original value, thus

Ln [A]t/[A]₀ = Ln 1/2[A]₀/[A]₀= Ln 1/2 = - kt

- 0.693 = - k(530s) ⇒ k = 1.3 x 10⁻³ s⁻¹

4 0

2 years ago

You decide to establish a new temperature scale on which the melting point of ammonia (-77.75 ∘c) is 0∘a, and the boiling point

Hoochie [10]

First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:

-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
x = 177.75*100/44.4 = 400.33

The boiling point of water in ∘a would be 400.33∘a.

7 0

2 years ago

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2CH4(g)⟶C2H4(g)+2H2(g)

Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

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According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of $CH_4$ will be,