All categories

3 years ago

Please help me with this question

Chemistry

1 answer:

ollegr [7]3 years ago

6 0

Answer:

It shows constant motion when a straight line( not parallel) is formed in distance time graph.

You might be interested in

A 2.50 L balloon is filled with water at 2.27 atm. If the balloon is squeezed into a 0.80 L beaker and does NOT burst, what is t

Ludmilka [50]

Answer:

$P_2=7.09atm$

Explanation:

Hello!

In this case, given the change in volume and pressure of the gas, it is possible for us to recall the Boyle's law as way to understand the inversely proportional relationship between pressure and volume:

$P_1V_1=P_2V_2$

Thus, when solving for the final pressure, P2, given the initial pressure and volume and the final volume, we obtain:

$P_2=\frac{P_1V_1}{V_2}\\\\P_2=\frac{2.27atm*2.50L}{0.80L}\\\\P_2=7.09atm$

Best regards!

3 0

3 years ago

Read 2 more answers

Choose the words that make each statement true.

Ksenya-84 [330]

Answer:

moles, liter, concentration, and 6.0 mol/L

Explanation:

3 0

4 years ago

Measurements show that unknown compound has the following composition: element mass 62.1 % carbon, 10.5 % hydrogen and 27.6 % ox

tino4ka555 [31]

Answer:

C3H6O

Explanation:

The percentage composition of the elements in the compound are given as follows:

62.1 % carbon = 62.1g of C

10.5 % hydrogen = 10.5g of H

27.6 % oxygen = 27.6g of O

Next, we convert each mass to mole by dividing by their molar/atomic mass

C = 62.1/12 = 5.175mol

H = 10.5/1 = 10.5mol

O = 27.6/16 = 1.725mol

Next, we divide each mole value by the smallest mole value (1.725)

C = 5.175mol ÷ 1.725 = 3

H = 10.5mol ÷ 1.725 = 6.086

O = 1.725mol ÷ 1.725 = 1

The empirical ratio approximately of C:H:O is 3:6:1, hence, the empirical formula is C3H6O

5 0

3 years ago

2. The table below shows the atomic numbers of elements of the periodic table represented by letters I to Q The letters are not

Komok [63]

Explanation:

cudhdbsieehe eirbruwnee8ebeuw eodjsuww8

4 0

2 years ago

The rate constant of the first-order reaction A→3B is 0.291 s−1. The concentration of A at t=5 seconds is 0.081 mol/L. What was

zimovet [89]

Answer : The initial concentration of A is, 0.347 mol/L

Explanation :

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $0.291s^{-1}$

t = time passed by the sample = 5 s

a = initial amount of the reactant = ?

a - x = amount left after decay process = 0.081 mol/L

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

$5=\frac{2.303}{0.291}\log\frac{a}{0.081}$

$a=0.347mol/L$

Therefore, the initial concentration of A is, 0.347 mol/L

7 0

3 years ago