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3 years ago

Choose all that apply. Which of the following are included in the physical sciences? biology sociology physics geology

Chemistry

2 answers:

Jet001 [13]3 years ago

6 0

Hi there.

The correct answers are Physics and Geology

Bond [772]3 years ago

3 0

Physics and biology are the correct answers

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Aluminum chloride is added to cesium does a single replacement reaction occur, and if so what is the correct chemical equation f

krek1111 [17]

The answer is C

I hope that helped

5 0

2 years ago

Read 2 more answers

Determine the approximate amount of potassium hydrogen phthalate, KHP, that you will need to neutralize 6.00 ml of 0.100 M NaOH.

Sveta_85 [38]

Answer:

potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol

to get 1000 ml

Molar concentration = Mass concentration/Molar Mass

mass concentration = molar concentration x molar mass

mass concentration=0.1 M,

molar mass= 204.233 g/mol

so to get 1L

mass conc = 204.233 x 0.1

= 20.4233g for 1L or 1000 ml

to get 6.00 ml

if 20.4233g is for 1000ml

then to 6.00 ml

= 20.4233 x 6 / 1000

= 0.123g for 6.00 ml

according to the equation below

NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)

number of moles of NaOH is equal to that of KHP

so the same amount will be needed too, which is

= 0.123g

6 0

2 years ago

Need help balancing equations please.

densk [106]

Answer:

A, 4,3,2

B, 1,1,2

C, 1,1,1,2

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2 years ago

You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30

dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula: $\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n$

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{1}{3})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = 0.012345$

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

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3 years ago

The enzyme urease catalyzes the breakdown of urea in the body. Urease breaks urea down to 2NH3+CO2. This is an example of a hydr

Alika [10]

Answer:

Look at the picture.

Explanation:

On stage one binding of a substrate occurs (and also the geometry of active site may change) and water comes to the site. On stage two the hydrolisis takes place and on stage 3 products deabsorb from the enzyme.

8 0

2 years ago