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Helen [10]
3 years ago
5

The new hybrid car can get 51.5 km/gal. It has a top speed of 40000.00 cm/min and is 4m long. How fast can the car go in m/hr?

Chemistry
1 answer:
horrorfan [7]3 years ago
4 0

Answer:

The anawer of this question is 0.024 m/h

Explanation:

Other explanations of the question are additional.

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What is the molarity of a solution made by mixing 50.0 g of magnesium nitrate, Mg(NO3)2, in enough water to make 250. mL of solu
Ira Lisetskai [31]
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A solution
Gelneren [198K]
A solution <span>has a uniform composition and is only able to be separated by chemical means.</span>
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(b) Data has been collected to show that at a given wavelength in a 1 cm pathlength cell, Beer's Law for the absorbance of Co2+
OlgaM077 [116]

Answer : The concentration of a solution with an absorbance of 0.460 is, 0.177 M

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution

l = path length

\epsilon = molar absorptivity coefficient

From this we conclude that absorbance of solution is directly proportional to the concentration of solution at constant path length.

Thus, the relation between absorbance and concentration of solution will be:

\frac{A_1}{A_2}=\frac{C_1}{C_2}

Given:

A_1 = 0.350

A_2 = 0.460

C_1 = 0.135 M

C_2 = ?

Now put all the given values in the above formula, we get:

\frac{0.350}{0.460}=\frac{0.135}{C_2}

C_1=0.177M

Therefore, the concentration of a solution with an absorbance of 0.460 is, 0.177 M

3 0
3 years ago
F a sample of butene (C4H8) that has a mass of 136.6 g is combusted in excess oxygen, what is the mass of CO2 that is produced?
alukav5142 [94]

The balanced chemical reaction will be:

C4H8 + 6 O2 --> 4 CO2 + 4 H2O

We are given the amount of butene being combusted. This will be our starting point.

136.6 g C4H8 (1 mol  C4H8/ 56.11 g C4H8) (4 mol CO2/1 mol <span>C4H8</span>) ( 44.01 g CO2/ 1 mol CO2) = 428.6 g CO2
7 0
3 years ago
Read 2 more answers
Can somebody help me with this also plz
just olya [345]

ANSWER:

4 a) Specific elements have more than one oxidation state, demonstrating variable valency.

For example, the following transition metals demonstrate varied valence states: Fe^{2+}, Fe^{3+}, Cr^{2+}, Cr^{3+}, etc.

Normal metals such as Pb^{2+} and Pb^{4+} also show variable valencies. Certain non-metals are also found to show more than one valence state Pb^{3+} and Pb^{5+}.

4 b) Isotopes are members of a family of an element that all have the same number of protons but different numbers of neutrons.

For example, Carbon-14 is a naturally occurring radioactive isotope of carbon, having six protons and eight neutrons in the nucleus. However, C-14 does not last forever and there will come a time when it loses its extra neutrons and becomes Carbon-12.

5 a) 2Fe + 3Cl_2 → 2FeCl_3    

5 b) 3Pb + 8HNO_3 → 3Pb (NO_3)_2 + 4H_2O + 2NO_2

5 c) Zn + H_2SO_4 → ZnSO_4 + H_2   (already balanced so don't need to change)

5 d) 2H_2 + O_2 → 2H_2O

5 e) 2Mg + 2HCl → 2MgCl + H_2

EXPLANATION (IF NEEDED):

1. Write out how many atoms of each element is on the left (reactant side) and right (product side) of the arrow.

2. Start multiplying each side accordingly to try to get atoms of the elements on both sides equal.

EXAMPLE OF BALANCING:

8 0
2 years ago
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