Identify these element:a non-metal in the same group as leads?

coldgirl [10]

Answer:

$\frac{3}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{3 \sqrt{3} }{3} = \sqrt{3}$

when there is a radical in the denominator, we should rationalize (mutiply the denominator and numerator by the radical) to get rid of the radical in the denominator.

4 0

3 years ago

Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficien

Vadim26 [7]

Answer:

_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

7 0

3 years ago

What amount if force need to be applied to a 20 kg bowling ball to give it an acceleration of 5m/s

balandron [24]

Answer:

100N

Explanation:

Given parameters:

Mass of the bowling ball = 20kg

Acceleration = 5m/s²

Unknown:

Amount of force applied = ?

Solution:

To solve this problem, we apply newton's second law of motion.

Force = mass x acceleration

Now insert the parameters and solve;

Force = 20 x 5 = 100N

5 0

3 years ago

Consider the first-order reaction described by the equation At a certain temperature, the rate constant for this reaction is 5.8

zubka84 [21]

<u>Answer:</u> The half life of the reaction is 1190.7 seconds

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

k = rate constant of the reaction = $5.82\times 10^{-4}s^{-1}$

$t_{1/2}$ = half life of the reaction = ?

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{5.82\times 10^{-4}s^{-1}}=1190.7s$

Hence, the half life of the reaction is 1190.7 seconds

8 0

3 years ago

Why do you think large differences in temperature between the wet bulb and dry bulb thermometers are possible only at higher tem

olganol [36]

Answer:

This is the temperature indicated by a moistened thermometer bulb exposed to the air flow. The evaporation is reduced when the air contains more water vapor. The wet bulb temperature is always lower than the dry bulb temperature but will be identical with 100% relative humidity.

Explanation:

7 0

3 years ago