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Keith_Richards [23]
3 years ago
7

Bromine has chemical properties most similar to...

Chemistry
2 answers:
Margaret [11]3 years ago
6 0
It would be A. Fluorine
Aleksandr-060686 [28]3 years ago
3 0
Flourine.

You are very welcome
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What is catalyst? Write its types.​
Step2247 [10]

Answer:

A catalyst is a chemical substance that alters the rate of chemical reaction not consumed by the reaction. Hence, a catalyst can be recovered chen unchanged at the ends of chemical reaction. Catalyst can be divided into two typ the basis whether it speeds up or slowdowns the rate of chemical reaction. The positive catalyst and negative catalyst.

5 0
3 years ago
Read 2 more answers
If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
Our solar system formed from a
ankoles [38]
Our solar system formed about 4.5 billion years ago from a dense cloud of interstellar gas and dust. The cloud collapsed most likely due to a shockwave form a nearby exploding start called a supernova
8 0
3 years ago
In preparation for a blizzard, a city dispatches a crew to spread salt on the street surface of the bridges around town. The sal
Ilia_Sergeevich [38]
0 degrees to -6 or -16 degrees Celsius

3 0
3 years ago
If the distance between two objects is decreased to - of the original
Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

F = G\frac{m_1m_2}{R^2}The problem states  that  the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R

And the force at this distance would be written in terms of the same equation:

F_2 = G\frac{m_1m_2}{R_2^2}

Find the ratio between the final and the initial force:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}

Substitute the value for the final distance in terms of the initial distance:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}

Simplify:

\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}

This means the new force will be \frac{1}{100} of the original force.

8 0
3 years ago
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