All categories

4 years ago

Periodic TrendsPLEASE HELP ME! I WILL GIVE BRAINLIEST

Chemistry

1 answer:

xz_007 [3.2K]4 years ago

5 0

Answers:

A) Li

B) Ar

C) Br

D) Ne

E) B

You might be interested in

Is heat a from of kinetic or potential energy?

Leni [432]

Answer:

its made up of both but I would probably say kinetic

Explanation:

3 0

3 years ago

Read 2 more answers

Calculate the mass of 3.4 moles of nitric acid (hno3). explain the process or show your work by including all values used to det

balu736 [363]

Answer:

6.34917360^25g

Explanation:

It's been a while since I've done this type of problem so I'm not making any promises that its right hahaha, but I hope it helps anyway. Please let me know whether I'm right or not!

6 0

2 years ago

If further increases in substrate concentration do not result in further increases in reaction rate, then an enzyme is likely

igor_vitrenko [27]

Given what we know, we can confirm that if further increases in substrate concentration do not result in further increases in reaction rate, then an enzyme is likely saturated.

<h3>What does it mean for an enzyme to be saturated?</h3>

Enzymes work by binding to the substrate in specific zones of the enzyme. The zones are known as the active sites on enzymes. Since enzymes have a limited amount of these zones, once they are all bonded to a substrate, we can say that it is saturated.

Therefore, the saturation of enzymes allows us to explain how further increases in substrate concentration do not result in further increases in reaction rate.

To learn more about enzymes visit:

brainly.com/question/24811456?referrer=searchResults

4 0

2 years ago

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

natali 33 [55]

Answer:

There are 0.93 g of glucose in 100 mL of the final solution

Explanation:

In the first solution, the concentration of glucose (in g/L) is:

15.5 g / 0.100 L = 155 g/L

Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.

30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)

The concentration of the second solution is:

$155 \frac{g}{L} *\frac{0.030L}{0.500L}=9.3\frac{g}{L}$

So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:

1 L --------- 9.3 g

0.1 L--------- Xg

Xg = 9.3 g * 0.1 L / 1 L = 0.93 g

8 0

4 years ago

21. Solve the simultaneous equations graphically taking the values of

tigry1 [53]

Answer:

take the l my gang tfdfhngtyhggggggfggg

3 0

3 years ago

Periodic TrendsPLEASE HELP ME! I WILL GIVE BRAINLIEST​

Periodic TrendsPLEASE HELP ME! I WILL GIVE BRAINLIEST