All categories

3 years ago

ANSWER ASAP GIVING BRAINLIEST AND STUFF!

1 answer:

6 0

To find the average speed, you take the entire distance traveled (d) divided by the total time (t). (d)/(t)= average speed of object

You might be interested in

Hydrofluoric acid is what type of acid?

posledela

Hydrofluoric acid is a solution of hydrogen fluoride (HF) in water. Solutions of HF are colourless, acidic and highly corrosive. It is used to make most fluorine-containing compounds; examples include the commonly used pharmaceutical antidepressant medication fluoxetine (Prozac) and the material PTFE (Teflon).

5 0

2 years ago

Provide three different examples of how minerals can form on earth

Misha Larkins [42]

A mineral is a naturally occurring , solid, crystalline substance with a specific chemical composition. Minerals are usually inorganic and are formed by ionic, covalent or metallic bonding. Ionic bonds are the dominant type of bonds in mineral strictures. 90% of all minerals are ionic compounds. Minerals bonded by covalent bonds are strong, for example carbon bonded together to form diamond. Metallic bonds are a type of covalent bonds where the atoms have a strong tendency to lose electrons and pack together as cations.

6 0

3 years ago

Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

7O₂: 7 moles of 1 O=O bond

4CO₂: 4 moles of 2 C=O bonds

6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$

$\Delta H = -2860 kJ/mol$

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

I hope it helps you!

7 0

3 years ago

Drug testing of athletes and other individuals is big business.

ziro4ka [17]

Answer:

bioanalytical chemistry

Explanation:

7 0

3 years ago

A solution is made by dissolving 0.0150 mol of HF in enough water to make 1.00 L of solution. At 26 °C, the osmotic pressure of

Alex777 [14]

Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm

Required:
percent ionization

Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367

Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x

Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3

percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%

Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>

7 0

3 years ago