How many moles of oxygen are required in order to produce 4 moles of water?

If an unknown element didplays many extremly strong metallic properties, where is it most likely to be located on the periodic t

Ipatiy [6.2K]

This is late but for anyone else who needs it...It's D. Far left

4 0

3 years ago

A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d

Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

8 0

2 years ago

What is the concentration of a solution that has 15.0 g NaCl dissolved to a total of 750 ml?

Oksana_A [137]

Answer: The concentration of solution is 0.342 M

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of solute (Sodium chloride) = 15 g

Molar mass of sodium chloride = 58.5 g/mol

Volume of solution = 750 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{15g\times 1000}{58.5g/mol\times 750mL}\\\\\text{Molarity of solution}=0.342M$

Hence, the concentration of solution is 0.342 M

6 0

2 years ago

What do you mean by commercial farming

Zinaida [17]

Answer

Commercial farming is a large-scale production of crops for sale, intended for widespread distribution to wholesalers or retail outlets.

5 0

3 years ago

An ideal gas in a sealed container has an initial volume of 2.80 L. At constant pressure, it is cooled to 18.00 °C, where its

Aleks04 [339]

Answer:

$T_1=-91.18\°C$

Explanation:

Hello there!

In this case, given the T-V variation, we understand it is possible to apply the Charles' law as shown below:

$\frac{T_1}{V_1}= \frac{T_2}{V_2}$

Thus, since we are interested in the initial temperature, we can solve for T1, plug in the volumes and use T2 in kelvins:

$T_1= \frac{T_2V_1}{V_2}\\\\T_1= \frac{(18.00+273.15)K(1.75L)}{(2.80L)}\\\\T_1=182K-273.15\\\\T_1=-91.18\°C$

Best regards!

6 0

3 years ago