What have you observed when you fill in a basin with water? how do you do it?​

Answers

Answer 1
Answer:

When you fill a basin with liquid water, you can see that the water takes the shape of the container in which it is contained. This is because in the liquid state, water has molecules farther apart than in the solid state.

You can notice this property when performing an experiment with liquid and solid water.

When filling a glass, liquid water takes on the shape of a glass, and solid water, such as an ice cube, remains the same shape when placed in a glass.

Therefore, when filling a basin with water we perceive a property of the physicalstate of water, in liquid form. Water is one of the few substances that can be found naturally in liquid, solid and gaseous states.

Learn more here:

brainly.com/question/23650420

Answer 2
Answer:

Answer:

Cautiously and avoiding filling in the central area so that it does not overflow when filling, since being very beach makes filling difficult.

Explanation:

The basins are shallow, that is why filling is difficult, the filling must be slow, low intensity and at the edges not placing the water filling in the center of the basin.


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Use the following structures of amino acids to answer the questions below. Note that the difference in the structures (the side chains) is highlighted by gray shading.


A student performed chromatography of the four amino acids and theresults were shown in the chromatogram below. If an anion exchangecolumn (column is positively charged) was used in a neutral buffer,assign each amino acid to the corresponding peak in the chromatogram.

Answers

In a positively charged column, Asparate will travel the farthest followed by Threonine, Leucine and Lysine.

Which statement is true about the cell membrane? It allows an unlimited amount of water to enter. It stops waste substances from passing through. It is not selective about which gases enter or leave the cell. It is selective about which substances enter or leave the cell.

Answers

Answer:

It is selective about which substances enter or leave the cell.

Explanation:

Also srry for spamming u all the time.

UwU hope this helps tho

Answer:

It is selective about which substances enter or leave the cell

Explanation:

I have learned this already

Balance the chemical equation given below, and determine the number of moles of iodine that reacts with 10.0 g of aluminum._____ Al(s) + _____ I2(s) → _____ Al2I6(s)

Answers

Answer:

1. 2Al + 3I2 —> Al2I6

2. 0.555mol of I2

Explanation:

1. Al + I2 —> Al2I6

Observing the above equation, there are 2 atoms of Al on the right side and 1 on the left side. To balance it, put 2 in front of Al as shown below:

2Al + I2 —>Al2I6

Also, there are 6 atoms of I on the right side and 2 on the left side. To balance it, put 3 in front I2 as shown below:

2Al + 3I2 —>Al2I6

2. Molar Mass of Al = 27g/mol

Mass of Al = 10g

n = Mass /Molar Mass

n = 10/27 = 0.37mol

From the equation,

2moles of Al reacted with 3 moles of I2.

Therefore, 0.37mol of Al will react with = (0.37 x 3)/2 = 0.555mol of I2

The balanced chemical equation is:

2 Al(s) + 3 I₂(s) → Al₂I₆(s)

0.557 moles of iodine react with 10.0 g of aluminum.

Let's consider the following unbalanced equation.

Al(s) + I₂(s) → Al₂I₆(s)

We will balance it using the trial and error method. We can get the balanced equation by multiplying Al by 2 (balance Al atoms) and I₂ by 3 (balance I atoms).

2 Al(s) + 3 I₂(s) → Al₂I₆(s)

The molar mass of Al is 26.98 g/mol. The moles corresponding to 10.0 g of Al are:

10.0 g * (1mol)/(26.98g) = 0.371 mol

The molar ratio of Al to I₂ is 2:3. The moles of I₂ that react with 0.371 moles of Al are:

0.371 molAl * (3molI_2)/(2molAl) = 0.557 mol I_2

The balanced chemical equation is:

2 Al(s) + 3 I₂(s) → Al₂I₆(s)

0.557 moles of iodine react with 10.0 g of aluminum.

You can learn more about stoichiometry here: brainly.com/question/9743981

Calculate the number of C atoms in 9.837 x 1024 molecules of CO2.

Please help

Answers

Answer:

Explanation:

1 molecule contains 1 carbon atom.

9.837 * 10^24 molecules contains 9.837 * 10^24  atom of carbon.

It's a 1 to 1 ratio.

Which term describes this molecular shape?A. Trigonal pyramidal
B. Trigonal planar
C. Tetrahedral
D. Linear

Answers

Answer: Trigonal Pyrimidal

Explanation: Just took test

The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen, if the rate of the formation of carbon monoxide is 0.35 M/s ? g

Answers

Answer :  The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-(1)/(a)(d[A])/(dt)

\text{Rate of disappearance of B}=-(1)/(b)(d[B])/(dt)

\text{Rate of formation of C}=+(1)/(c)(d[C])/(dt)

\text{Rate of formation of D}=+(1)/(d)(d[D])/(dt)

Rate=-(1)/(a)(d[A])/(dt)=-(1)/(b)(d[B])/(dt)=+(1)/(c)(d[C])/(dt)=+(1)/(d)(d[D])/(dt)

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

CH_4+H_2O\rightarrow CO+3H_2

The expression for rate of reaction :

\text{Rate of disappearance of }CH_4=-(d[CH_4])/(dt)

\text{Rate of disappearance of }H_2O=-(d[H_2O])/(dt)

\text{Rate of formation of }CO=+(d[CO])/(dt)

\text{Rate of formation of }H_2=+(1)/(3)(d[H_2])/(dt)

The rate of reaction expression is:

\text{Rate of reaction}=-(d[CH_4])/(dt)=-(d[H_2O])/(dt)=+(d[CO])/(dt)=+(1)/(3)(d[H_2])/(dt)

As we are given that:

+(d[CO])/(dt)=0.35M/s

Now we to determine the rate for the formation of hydrogen.

+(1)/(3)(d[H_2])/(dt)=+(d[CO])/(dt)

+(1)/(3)(d[H_2])/(dt)=0.35M/s

(d[H_2])/(dt)=3* 0.35M/s

(d[H_2])/(dt)=1.05M/s

Thus, the rate for the formation of hydrogen is, 1.05 M/s