C. Neither a Physical or Chemical Changu

Hands moving on a battery-operated clock is an example of what kind of

Alexxandr [17]

The C is right answer

4 0

2 years ago

An unknown compound has a percent composition of 52.10% potassium, 15.8% carbon, and 32.1% oxygen. The molar mass of the compoun

lianna [129]

To determine the empirical formula and the molecular formula of the compound, we assume a basis of the compound of 100 g. We do as follows:

Mass Moles
K 52.10 52.10/39.10 = 1.33 1.33/1.32 ≈ 1
C 15.8 15.8/12 = 1.32 1.32/1.32 ≈ 1
O 32.1 32.1 / 16 = 2.01 2.01/1.32 ≈ 1.5

The empirical formula would most likely be KCO.
The molecular formula would be K2C2O3.

4 0

3 years ago

Read 2 more answers

Give an example of how environmental science might involve geology and chemistry.

Sever21 [200]

One of the examples is radiation and chemistry of water. Environmental science requires the capacity to integrate data from the greater part of the significant fields of science, and in addition from arithmetic.
Geology is vital on the grounds that huge scale arrives forms make geology. The presence of mountains and valleys influences how much daylight and precipitation achieve the ground, how breezy an area is, the manner by which precipitation keeps running off, and numerous different variables that figure out what plants and creatures will have the capacity to occupy a district.

8 0

2 years ago

a culinary student fills a 40ml container with 37.2 grams of vegetable oil.What is the density of the oil

Lynna [10]

Answer:

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 37.2 g

volume = 40 mL

We have

$density = \frac{37.2}{40} \\ = 0.93$

We have the final answer as

Hope this helps you

4 0

2 years ago

A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________

gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

R = gas constant, T = temperature, Mm = molar mass of the gas particles

From the question

R = 8,314 J / mol K

T = temperature

Mm = molar mass, kg / mol

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

$\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s$

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

M₁ = molar mass sulfur dioxide = 64

M₂ = molar mass nitrogen triodide = 395

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5$

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0

2 years ago