All categories

3 years ago

What is the name of the element? bromine iodine tellurium xenon

Chemistry

2 answers:

zhannawk [14.2K]3 years ago

4 0

Answer:

iodine.

guajiro [1.7K]3 years ago

4 0

Answer:

iodine

Explanation:

You might be interested in

Ca(hco3)2 molecular formula calculated

GrogVix [38]

Answer:

C2H2CaO6

Explanation:

Ca same

H × 2

C × 2

O3 × 2

I hope this helps you :)

7 0

3 years ago

Resistance is the slowing of the flow of electrons and where some of the electrical energy is converted into heat. true or fals

zaharov [31]

Answer: True

Explanation:

I just took the test and got it right

3 0

3 years ago

What is the wavelength of a wave having a frequency of 2500 Hz

Lemur [1.5K]

We will assume that the wave is moving in free space, thus, the velocity of the wave would be equal to the speed of light = 3 * 10^8 meters/sec

The velocity of the wave can be calculated using the following rule:
velocity = frequency * wavelength where:
velocity = 3 * 10^8 m/sec
frequency = 2500 Hz
wavelength is the unknown that we want to calculate.

Substitute with the givens in the above equation to get the wavelength as follows:
3 * 10^8 = 2500*wavelength
wavelength = 120 * 10^3 meters

5 0

3 years ago

Read 2 more answers

At a particular temperature a 2.00-L flask at equilibrium contains 2.80 ✕ 10-4 mol N2, 2.50 ✕ 10-5 mol O2, and 2.00 ✕ 10-2 mol N

zhenek [66]

Answer : The value of equilibrium constant (K) is, $5.71\times 10^4$

Explanation :

First we have to calculate the concentration of $N_2,O_2\text{ and }N_2O$

$\text{Concentration of }N_2=\frac{\text{Moles of }N_2}{\text{Volume of solution}}=\frac{2.80\times 10^{-4}mol}{2.00L}=1.4\times 10^{-4}M$

and,

$\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{2.50\times 10^{-5}mol}{2.00L}=1.25\times 10^{-5}M$

and,

$\text{Concentration of }N_2O=\frac{\text{Moles of }N_2O}{\text{Volume of solution}}=\frac{2.00\times 10^{-2}mol}{2.00L}=1.00\times 10^{-2}M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

$N_2(g)+O_2(g)\rightarrow 2N_2O(g)$

The expression for equilibrium constant is:

$K=\frac{[N_2O]^2}{[N_2][O_2]}$

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

$K=\frac{(1.00\times 10^{-2})^2}{(1.4\times 10^{-4})\times (1.25\times 10^{-5})}$

$K=5.71\times 10^4$

Thus, the value of equilibrium constant (K) is, $5.71\times 10^4$

7 0

4 years ago

Wastewater from a cement factory contains 0.280 g of Ca2+ ion and 0.0220 g of Mg2+ ion per 100.0 L of solution. The solution den

faltersainse [42]

<u>Answer:</u> The concentration of $Ca^{2+}\text{ and }Mg^{2+}$ ions are 2.797 ppm and 0.212 ppm respectively.

<u>Explanation:</u>

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of gold = 100 L = 100000 mL (Conversion factor: 1 L = 1000 mL)

Density of gold = 1.001 g/mL

Putting values in above equation, we get:

$1.001g/mL=\frac{\text{Mass of solution}}{100000mL}\\\\\text{Mass of solution}=1.001\times 10^5g$

To calculate the concentration in ppm (by mass), we use the equation:

$ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

<u>Calculating the concentration of calcium ions:</u>

Mass of $Ca^{2+$ ions = 0.280 g

Putting values in above equation, we get:

$ppm(Ca^{2+})=\frac{0.280g}{1.001\times 10^5}\times 10^6=2.797ppm$

<u>Calculating the concentration of magnesium ions:</u>

Mass of $Mg^{2+$ ions = 0.0220 g

Putting values in above equation, we get:

$ppm(Mg^{2+})=\frac{0.0220g}{1.001\times 10^5}\times 10^6=0.212ppm$

Hence, the concentration of $Ca^{2+}\text{ and }Mg^{2+}$ ions are 2.797 ppm and 0.212 ppm respectively.

7 0

3 years ago