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Aleonysh [2.5K]
3 years ago
7

4. The [H+] in solution is 1.0 X 10-3 M. This solution is BEST classified as?

Chemistry
2 answers:
Afina-wow [57]3 years ago
6 0
I believe the answer is c
Dennis_Churaev [7]3 years ago
3 0
I would be neutral because h20 has a impact on water
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The gas in a balloon occupies 2.25 L at 298 K and 300 kPa. At
o-na [289]

If a gas in a balloon occupies 2.25 L at 298 K and 300 kPa, the temperature at which the balloon expand to 3.50L and 2.17 atm is 345.23K.

<h3>How to calculate temperature?</h3>

The temperature of an ideal gas can be calculated using the following formula:

P1V1/T1 = P2V2/T2

Where;

  • P1 = initial pressure
  • P2 = final pressure
  • V1 = initial volume
  • V2 = final volume
  • T1 = initial temperature
  • T2 = final temperature

According to the information given in this question;

  • P1 = 300kpa = 2.96 atm
  • P2 = 2.17 atm
  • V1 = 2.25L
  • V2 = 3.50L
  • T1 = 298K
  • T2 = ?

2.96 × 2.25/298 = 2.17 × 3.5/T2

0.022T2 = 7.595

T2 = 7.595 ÷ 0.022

T2 = 345.23K

Therefore, if a gas in a balloon occupies 2.25 L at 298 K and 300 kPa. the temperature at which the balloon expand to 3.50L and 2.17 atm is 345.23K.

Learn more about temperature at: brainly.com/question/11464844

5 0
2 years ago
How many moles would be in 85.OmL of 0.750M KOH?
Harlamova29_29 [7]

Answer: There are 0.0637 moles present in 85.0 mL of 0.750 M KOH.

Explanation:

Given: Volume = 85.0 mL (1 mL = 0.001 L) = 0.085 L

Molarity = 0.750 M

It is known that molarity is the number of moles of solute present in liter of a solution.

Therefore, moles present in given solution are calculated as follows.

Molarity = \frac{moles}{Volume (in L)}\\0.750 M = \frac{moles}{0.085 L}\\moles = 0.0637 mol

Thus, we can conclude that there are 0.0637 moles present in 85.0 mL of 0.750 M KOH.

7 0
3 years ago
A chemist is given a sample of the CuSO4 hydrate and asked to determine its the empirical formula. The original sample weighed 4
Firlakuza [10]

Answer: The formula for this hydrate is CuSO_4.5H_2O

Explanation:

Decomposition of hydrated copper sulphate is given by:

CuSO_4.xH_2O\rightarrow CuSO_4+xH_2O

Molar mass of CuSO_4 = 160 g/mol

According to stoichiometry:

(160+18x) g of CuSO_4.xH_2O decomposes to give 160 g of anhydrous CuSO_4

Thus 42.75  g of CuSO_4.xH_2O decomposes to give=\frac{160}{160+18x}\times 42.75g of anhydrous CuSO_4

\frac{160}{160+18x}\times 42.75=27.38

x=5

Thus the formula for this hydrate is CuSO_4.5H_2O

8 0
3 years ago
In this experiment, you will use a natural indicator to indirectly determine the H+ concentration of some common household solut
valentinak56 [21]

Some of the scientific questions that may be answered through the experiment are:

(1) What are the physical changes that may occur in the solution or the indicator when added with acidic/basic solution?

(2) How much of the indicator is needed in order to bring about a significant physical change in the solution to identify its H+ concentration?

if wrong i sry :(

4 0
3 years ago
Read 2 more answers
HCl, HC2H3O2, and H2SO4 are examples of
irina1246 [14]
First of all, it must be noted that all these substances presented above are examples of compounds it that they are made of one or more elements that are present in fixed ratios. Then, it may also be noted that all of them contain H (hydrogen), this type of substances are ACIDS. 
5 0
3 years ago
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