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Dmitriy789 [7]
3 years ago
13

Low tide____.

Chemistry
2 answers:
Fantom [35]3 years ago
8 0

Answer:

i think the answer is A

Explanation:

because in some areas, a regular pattern occurs of one high tide and one low tide each day,

marin [14]3 years ago
6 0

Answer:

A) happens once a day.

Explanation:

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Consider the following gas phase reaction:
snow_tiger [21]

Answer:

2NO(g) + O2(g) --> 2NO2(g)

now 400 ml of NO × 2 mol of NO2/2 mol of NO

= 400 ml of NO2

now 500 ml of O2 × 2 mol of NO2/1 mol of O2

= 1000 ml of NO2

now 400 ml of NO2 × 1 mol of O2/2 mol of NO

= 200 ml

subtract that from 500 ml of total i.e. 500-200 =300 ml

The total volume of the reaction mixture is 1000 ml -300ml = 700 ml

6 0
3 years ago
Calculate the molecular weight when a gas at 25.0 ∘C and 752 mmHg has a density of 1.053 g/L . Express your answer using three s
stiks02 [169]

Answer:

26.0 g/mol is the molar mass of the gas

Explanation:

We have to combine density data with the Ideal Gases Law equation to solve this:

P . V = n . R .T

Let's convert the pressure mmHg to atm by a rule of three:

760 mmHg ____ 1 atm

752 mmHg ____ (752 . 1)/760 =  0.989 atm

In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.

Moles = Mass / molar mass.

We can replace density data as this in the equation:

0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K

(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x

0.0405 mol = 1.053 g / x

x =  1.053 g / 0.0405  mol = 26 g/mol

7 0
4 years ago
Substrate: N-benzoyl-L-tyrosine ethyl ester (BTEE) 0.001M
Anna71 [15]

Based on the dilution formula, 0.1 mL of the stock solution of the enzyme is required to prepare a 50-fold diluted enzyme in 0.01 M HCl.

<h3>How can 50-fold dilution of the enzyme be done?</h3>

The 50-fold dilution of the stock enzyme solution can be done by using the dilution formula to determine the given volume of the stock solution required.

The dilution formula is given below:

  • C1V1 = C2V2

where:

  • C1 = Initial concentration of enzyme
  • C2 = Final concentration of enzyme
  • V1 = Initial volume
  • V2 = Final volume

From the given data for the enzyme dilution;

C1 = 1

C2 = 1/50 = 0.02

V1 = x

V2 = 5 ml

Making V1 subject of formula in the dilution formula:

V1 = C2V2/C1

V1 =  0.02 * 5/1 = 0.1 mL

Therefore, 0.1 mL of the stock solution of the enzyme is required to prepare a 50-fold diluted enzyme in 0.01 M HCl.

Learn more about dilution at: brainly.com/question/24709069

#SPJ1

7 0
2 years ago
2. Find the empirical formula for a compound which contains 32.8% chromium and 67.2%<br> chlorine.
Blababa [14]

The empirical formula is CrCl₃.

Explanation:

To find the empirical formula we use the following algorithm.

We divide each percentage by the atomic weight of each element:

for Cr   32.8 / 52 = 0.63

for Cl   67.2 / 35.5 = 1.89

The result will be divined by the smallest number:

for Cr    0.63 / 0.63 = 1

for Cl     1.89 / 0.63 = 3

The empirical formula is CrCl₃.

Learn more about:

empirical formula

brainly.com/question/5834630

brainly.com/question/12973081

#learnwithBrainly

6 0
4 years ago
How much heat energy would be needed to raise the temperature of a 223 g sample of aluminum [(C=0.895 Jig Cy from 22.5°C to 55 0
dsp73

Answer : The heat energy needed would be, 6486.5125 J

Explanation :

To calculate the change in temperature, we use the equation:

q=mc\Delta T\\\\q=mc(T_2-T_1)

where,

q = heat needed = ?

m = mass of aluminum = 223 g

c = specific heat capacity of aluminum = 0.895J/g^oC

\Delta T = change in temperature

T_1 = initial temperature = 22.5^oC

T_2 = final temperature = 55.0^oC

Putting values in above equation, we get:

q=223g\times 0.895J/g^oC\times (55.0-22.5)^oC

q=6486.5125J

Therefore, the heat energy needed would be, 6486.5125 J

5 0
3 years ago
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