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ollegr [7]
3 years ago
7

5TH GRADE SCIENCE!!!!!!

Chemistry
2 answers:
Yuri [45]3 years ago
4 0

Answer:

Phase changes that require a loss in energy are condensation and freezing.

Explanation:

nika2105 [10]3 years ago
3 0

Answer:

freezing consensation

Explanation:

You might be interested in
At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements
Gnom [1K]

Given that:

  • At 25C the density of water is 0.997044 g/mL.

From the information attached below, we have the following parameters.

The density of water calculation using a bottle.

     Initial volume of    Final volume of    Mass of water   Density (g/mL)

     burette (mL)        burette   (mL)       dispensed (g)

 

Sample 1      2.33                     7.34                   5.000               -----

Sample 2      7.34                    12.37                 5.025                -----

Sample 3      12.37                   18.50                6.112                  -----

Sample 4      18.50                  24.57               6.064                 -----

Sample 5     24.57                  31.31                6.720                  -----

The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.

  • The change in the volume of the burette = (final volume - the initial volume) mL

Sample 1:

= (7.34 - 2.33) mL

= 5.01 mL

Sample 2:

= (12.37 - 7.34) mL

= 5.03 mL

Sample 3:

= (18.50 - 12.37) mL

= 6.03 mL

Sample 4:

= (24.57 - 18.50) mL

= 6.07 mL

Sample 5:

= (31.31 - 24.57) mL

= 6.74 mL

The mass of the water dispersed in sample 1 is given as = 5.000 g

Using the relation for calculating the density of each, we have:

Sample 1

\mathbf{density = \dfrac{mass}{volume}}

\mathbf{density = \dfrac{5.01 g}{5.000 ml}}

density = 0.998004 g/ml

Sample 2:

\mathbf{density = \dfrac{5.025 g}{5.03ml}}

density = 0.999006 g/ml

Sample 3:

\mathbf{density = \dfrac{6.112 g}{6.13ml}}

density = 0.997064 g/ml

Sample 4:

\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}

density = 0.999012 g/ml

Sample 5:

\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}

density = 0.997033 g/ml

Thus, the average density for all the samples is:

\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 +   0.999012  + 0.997033  )}{5}}

= 0.998024

∴

The percentage error for the two densities measurement is:

=\dfrac{ (experimental \  value -theoretical  \ value)\times 100 }{theoretical  \ value}

Given that the theoretical value = 0.997044 g/ml

Then;

\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

Learn more about density here:

brainly.com/question/24386693?referrer=searchResults

4 0
2 years ago
What is the ph of 0.450 m al(no3)3 [ka for al3+(aq) = 1.00x10-5]? express your answer to two decimal places?
White raven [17]

Answer : The pH of the solution is, 2.67

Explanation :

The equilibrium chemical reaction is:

                           Al^{3+}+H_2O\rightarrow Al(OH)^{2+}+H^+

Initial conc.       0.450                   0               0

At eqm.           (0.450-x)                 x               x

As we are given:

K_a=1.00\times 10^{-5}

The expression for equilibrium constant is:

K_a=\frac{(x)\times (x)}{(0.450-x)}

Now put all the given values in this expression, we get:

1.00\times 10^{-5}=\frac{(x)\times (x)}{(0.450-x)}

x=0.00212M

The concentration of H^+ = x = 0.00212 M

Now we have to calculate the pH of solution.

pH=-\log [H^+]

pH=-\log (0.00212)

pH=2.67

Therefore, the pH of the solution is, 2.67

8 0
3 years ago
If I initially have a gas at a pressure of 12 kPa, a volume of 23 liters, and a temperature of 200 K, and then I raise the press
sertanlavr [38]

Answer:

The new volume of gas would be 30 L.

Explanation:

This is an example of a Combined Gas Laws problem.

5 0
3 years ago
How much ch2o is needed to prepare 445 ml of a 2.65 m solution of ch2o?
worty [1.4K]

Answer: 35.4 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

Molality = 2.65

n= moles of solute =?

 V_s = volume of solution in ml = 445 ml

Putting in the values we get:

2.65=\frac{n\times 1000}{445ml}

n=1.18

Mass of solute in g=moles\times {\text {molar mass}}=1.18mol\times 30.02g/mol=35.4g

Thus 35.4 grams of CH_2O is needed to prepare 445 ml of a 2.65 m solution of CH_2O.

8 0
2 years ago
What does AS> O mean?
Vlad1618 [11]

Answer:

ΔS> 0 means Letter A

Explanation:

Processes that involve an increase in entropy of the system (ΔS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:

\displaystyle \Delta {S}_{\text{univ}}=\Delta {S}_{\text{sys}}+\Delta {S}_{\text{surr}}

5 0
3 years ago
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