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melamori03 [73]
3 years ago
8

If a car travels 400.0 meters in 20.0 seconds, how fast is it going? m/s Please help me

Chemistry
1 answer:
Viefleur [7K]3 years ago
8 0

Answer:

20m/s.

Explanation:

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A _________ mixture is not well mixed.
Elza [17]
The answer would be D Heterogeneous
3 0
2 years ago
A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2
a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=13.00g

Mass of H_2O=2.662g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, \frac{12}{44}\times 13.00=3.54g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, \frac{2}{18}\times 2.662=0.296g of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = \frac{0.295}{0.295}=1

For Hydrogen = \frac{0.296}{0.295}=1

For Oxygen = \frac{0.603}{0.295}=2.044\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is CHO_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Hence, the empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

3 0
3 years ago
Which type of reaction occurs in the following equation?
Tema [17]
The  type  of  reaction  which  occurs  is  referred to  as  redox  reaction.  This  kind  of  reaction  involve  both  oxidation  and  reduction.
Al(s)  is  oxidized  to  alluminium  ions,  while  cu2+  is reduced  copper metal.Reduction  occurs   at  the  cathode  while oxidation  occurs at  the anode.

3 0
2 years ago
Read 2 more answers
Ou have a 5 ml sample of a protein in 0.5 m nacl. you place the protein/salt sample inside dialysis tubing (see fig. 2-14) and p
Oduvanchick [21]

Answer:

Procedure (2)  

Explanation:

Assume the dialyses come to equilibrium in the allotted times.

Procedure (1)

If you are dialyzing 5 mL of sample against 4 L of water, the concentration of NaCl will be decreased by a factor of

\dfrac{5}{4000} = \dfrac{1}{800}

Procedure (2)

For the first dialysis, the factor is

\dfrac{5}{1000} = \dfrac{1}{200}

After a second dialysis, the original concentration of NaCl will be reduced by a factor of  

\dfrac{1}{200} \times \dfrac{1}{200} = \dfrac{1}{40000}

Procedure (2) is more efficient by a factor of  

\dfrac{40000}{800} = \mathbf{50}

4 0
3 years ago
Complete the passage about ionic structures using words from the list
WARRIOR [948]

Answer:

see below

Explanation:

1- lattice

2- regular

3- positive

4- negative

5- alternate

6- strong

7- bonds

8- giant

7 0
2 years ago
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