Which comppund contains both ionic and covalent bonds

What does the arrow in chemical equations mean? (PLEASE HELP)

grin007 [14]

Answer:

yields

Explanation:

I just did that quiz on k12 8th grade

6 0

3 years ago

In this chemical formula: 3NH4 (ammonium) This chemical formula means there are 3 _____________ ___________, and 4 __________ __

seropon [69]

Answer:

hope it helped you

Explanation:

3 atoms of nitrogen and 4 atoms of hydrogen in each molecule

5 0

3 years ago

Depending upon the tip of a dropper pipet there are approximately 20 drops per milliliter of water. the experimental procedure p

Murrr4er [49]

Answer : The volume range in milliliters for the solution is 0.25 - 0.5 mL

Explanation :

We have been given that there are approximately 20 drops per milliliter of water

This information can be used as a conversion factor as $\frac{1mL}{20 drops}$

We are using 5-10 drops of the solution.

Let us calculate milliliters in 5 drops.

$5 drops \times\frac{1 mL}{20 drops} = 0.25 mL$

Similarly, 10 drops would contain

$10 drops \times\frac{1 mL}{20 drops} = 0.5 mL$ solution.

Therefore the volume range in milliliters for the solution is 0.25 - 0.5 mL

8 0

3 years ago

A container holds a sample of hydrogen gas at 300 K. The temperature decreased to 100 K and the pressure remained constant, and

Charra [1.4K]

Answer : The original volume of gas was, 3.0 L

Explanation :

Charles' Law : It is defined as the volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$

or,

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

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=?\\T_1=300K\\V_2=1.0L\\T_2=100K$

Now put all the given values in above equation, we get:

$\frac{V_1}{300K}=\frac{1.0L}{100K}\\\\V_1=3.0L$

Thus, the original volume of gas was, 3.0 L

8 0

3 years ago

A sample of carbon dioxide gas has a density of g/L at a pressure of 0.889 atm and a temperature of 55.0 °C. Assume ideal behavi

gladu [14]

Answer:

163.5 °C

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 686 mL.

Initial temperature (T1) = 55 °C.

Final volume (V2) = 913 mL

Initial pressure (P1) = final pressure (P2) = 0.889 atm

Final temperature (T2) =.?

Next, we shall convert celsius temperature to Kelvin temperature.

This can be done as shown below:

Temperature (K) = Temperature (°C) + 273

T(K) = T (°C) + 273

Initial temperature (T1) = 55 °C.

Initial temperature (T1) = 55 °C + 273

Initial temperature (T1) = 328 K

Next, we shall determine the new temperature of the gas.

Since the pressure is constant, we shall determine the new temperature as follow:

V1/T1 = V2 /T2

Initial volume (V1) = 686 mL.

Final volume (V2) = 913 mL

Initial temperature (T1) = 328 K

Final temperature (T2) =.?

V1/T1 = V2 /T2

686/328 = 913/T2

Cross multiply

686 x T2 = 328 x 913

Divide both side by 686

T2 = (328 x 913)/686

T2 = 436.5 K

Finally, we shall convert Kelvin temperature to celsius temperature.

This can be done as shown below:

Temperature (°C) = Temperature (K) – 273

T (°C) = T(K) – 273

T(K) = 436.5 K

T (°C) = 436.5 – 273

T (°C) = 163.5 °C

Therefore, the temperature of the gas sample is 163.5 °C.

5 0

3 years ago