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Harlamova29_29 [7]
3 years ago
13

It takes 38mL of 0.75M NaOH solution to completely neutralize 155 mL of a chloric acid solution (HclO3). what is the concentrati

on of the HclO3 solution? Please explain well
Chemistry
1 answer:
Scrat [10]3 years ago
8 0
ITS ABOUT DRIVE
ITS ABOUT POWER
WE STAY HUNGRY
WE DEVOUR
PUT IN THE WORK
PUT IN THE HOURS ⌚
AND TAKE WHATS OURS
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How can erode soil affect the body of water. ASAP​
Murrr4er [49]
Soil erosion has consequences that go beyond the loss of fertile land. It has resulted in an increase in contamination and sedimentation in lakes and rivers, clogging them and causing fish and other animals to decline. Degraded lands are also less capable of retaining water, which can exacerbate flooding.
6 0
3 years ago
Which of the following is composed of many cells?
vodomira [7]

Answer:

Bacteria

Explanation:

8 0
3 years ago
Question 22 of 30
Black_prince [1.1K]

Answer:

C6H12O6+6O2--->6CO2+6H2O

Explanation:

So I went through all the answers and could not find the right one amongst. If I'm not wrong the reaction above is the reaction for respiration. The nearest answer is D but unfortunately the first reactant isn't in accordance with that which the question has given.

8 0
3 years ago
A scuba diver knows that she needs 50.0mol of air for an upcoming dive. What size (volume) tank will she need to fill for this d
erica [24]

Answer:

1120 L.

Explanation:

Hello!

In this case, as no conditions of pressure of temperature are given for this problem, we can assume that the scuba diver dives at STP (1 atm and 273.15 K), which means that 1 mole of air would occupy a volume of 22.4 L.

In such a way, since she needs 50.0 moles of air, the following ratio is useful to compute the size (volume) of the tank she needs:

V_2=\frac{V_1*n_2}{n_1}

Thereby, we plug in to obtain:

V_2=\frac{22.4L*50.0mol}{1mol}\\\\V_2=1120 L

Best regards!

4 0
3 years ago
Given the following reaction: NH4SH (s) <--> NH3 (g) + H2S (g) If we start
almond37 [142]

Answer:

D. 0.3 M

Explanation:

                                              NH4SH (s)      <-->            NH3 (g) + H2S (g)

Initial concentration              0.085mol/0.25L             0                 0

Change in concentration     -0.2M                               +0.2 M        +0.2M

Equilibrium               0.035mol/0.25 L=0.14M             0.2M           0.2M

concentration

Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M

K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M

4 0
3 years ago
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