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2 years ago

Calculate how many moles of element Q are in 23.53 g of element Q.

Chemistry

1 answer:

pogonyaev2 years ago

7 0

Answer:

0.579 moles

Explanation:

Moles = 23.53/40.64

= 0.5789 moles

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It would have a solubility substance and surface

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3 years ago

A slender uniform rod 100.00 cm long is used as a meter stick. two parallel axes that are perpendicular to the rod are considere

kvv77 [185]

Refer to the diagram shown below.

The second axis is at the centroid of the rod.
The length of the rod is L = 100 cm = 1 m

The first axis is located at 20 cm = 0.2 m from the centroid.
Let m = the mass of the rod.

The moment of inertia about the centroid (the 2nd axis) is
$I_{g} = \frac{mL^{2}}{12} = (m \, kg) \frac{(1 \, m)^{2}}{12} = \frac{m}{12} \, kg-m^{2}$

According to the parallel axis theorem, the moment of inertia about the first axis is
$I_{1} = I_{g} + (m \, kg)(0.2 m)^{2} \\ I_{1} = \frac{m}{12}+ 0.04m = 0.1233m \, kg-m^{2}$

The ratio of the moment of inertia through the 2nd axis (centroid) to that through the 1st axis is
$\frac{I_{g}}{I_{1}} = \frac{0.0833m}{0.1233m} =0.6756$

Answer: 0.676

6 0

3 years ago

Read 2 more answers

Nh3 is a weak base (kb = 1.8 × 10–5 m) and so the salt nh4cl acts as a weak acid. what is the ph of a solution that is 0.013 m i

ankoles [38]

First, we need to get the value of Ka:

when Ka = Kw / Kb

we have Kb = 1.8 x 10^-5

and Kw = 3.99 x 10^-16 so, by substitution:

Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11

by using the ICE table :

NH4+ + H2O →NH3 + H+
intial 0.013 0 0

change -X +X +X

Equ (0.013-X) X X

when Ka = [NH3][H+] / [NH4+]

by substitution:

2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X

∴X = 5.35 x 10^-7

∴[H+] = X = 5.35 x 10^-7

∴PH = - ㏒[H+]

= -㏒(5.35 x 10^-7)
= 6.27

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3 years ago

Temperature in degrees Fahrenheit can be converted to Celsius by first subtracting 32, then dividing by 1.8. What is the Celsius

kap26 [50]

Answer:

31.1°C

Explanation:

Given parameters:

Temperature = 88°F

The formula of the to convert is:

T°F = T°C - 32 / 1.8 = $\frac{TC - 32}{1.8}$

Now input the parameters and solve;

T°F = $\frac{ 88 -32}{1.8}$

T°F = 31.1°C

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2 years ago

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If you stacked Avogadro's number of pennies one on top of the other on Earth's surface, how far would the stack extend (in kilom

attashe74 [19]

Answer:

Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km

<em>Note: The question is missing some parts. The complete question is as follows;</em>

<em>A penny has a thickness of approximately 1.0 mm . If you stack ed Avogadro's number of pennies one on top of the other on Earth 's surface, how far would the stack extend (in km)? [For comparison, the sun is about 150 million km from Earth and the nearest star (Proxim a Centauri) is about 40 trillion km from Earth.]</em>

Explanation:

Avogadro number = 6.02 * 10²³

thickness of a penny = 1.0 mm

I mm = 0.001 m

Thickness of Avogadro number of pennies stacked one upon another will be:

6.02 * 10²³ * 0.001 m = 6.02 * 10²⁰ m

Distance in km;

1 m = 0.001 km

therefore, 6.02 * 10²⁰ m = 6.02 * 10²⁰ * 0.001 km = 6.02 * 10¹⁷ km

Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km

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3 years ago