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mash [69]
3 years ago
14

1. Compound A contains 14 g of nitrogen for each 16 g of oxygen. Compound B contains 14 g of nitrogen for each 32 g of oxygen. W

hat is the lowest whole-number mass ratio of nitrogen that combines with a given mass of oxygen
Chemistry
1 answer:
Sever21 [200]3 years ago
4 0

Answer:

The lowest whole number mass ratio of nitrogen that combines with a given mass of oxygen is 1:1

Explanation:

In compound A, we have 14 g of nitrogen combine with  16 g of oxygen

In compound B, we have 14 g of nitrogen combine with  32 g of oxygen

Therefore,  we have

                                                                  A                            B

Ratio of molar masses                               14:16                 14:32

Mass of O that combine with 1 g of N       1.14                   2.29

Dividing by the smallest mass ratio          1                        2

Therefore the lowest whole number mass ratio at which oxygen and nitrogen can combine = 1:1.

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Which of the following is not equal to 120 centimeters?
tia_tia [17]
<h3>Answer:</h3>

0.012 dekameters (dkm)

<h3>Explanation:</h3>

<u>We are given;</u>

  • 120 centimeters( cm).

Required to identify the measurements that is not equivalent to 120 cm.

  • Centimeters are units that are used to measure length together with other units such as kilometers(km), meters (m), millimeters (mm), dekameters (dkm), etc.
  • These units can be inter-converted to one another using suitable conversion factors.
  • To do this, we are going to have a table showing the suitable conversion factor from one unit to another.

Kilometer (km)

10

Decimeter (Dm)

10

Hectometer (Hm)\

10

Meter (m)

10

Dekameter (dkm)

10

Centimeter (cm)

10

Millimeter (mm)

Therefore;

To convert cm to km

Conversion factor is 10^5 cm/km

Thus;

120 cm = 120 cm ÷ 10^5 cm/km

            = 0.0012 km

To convert cm to dkm

Conversion factor is 10 cm/dkm

Therefore,

120 cm = 120 cm ÷ 10 cm/dkm

            = 12 dkm

To convert cm to m

The suitable conversion factor is 10^2 cm/m

Thus,

120 cm = 120 cm ÷ 10^2 cm/m

           = 1.2 m

To convert cm to mm

Suitable conversion factor is 10 mm/cm

Therefore;

120 cm = 120 cm × 10 mm/cm

            = 1200 mm

Therefore, the measurement that is not equal to 120 cm is 0.012 dkm

4 0
3 years ago
A gas with a volume of 500. mL at 75°C is heated to 225 °C. What is the new volume of the gas?
mihalych1998 [28]
Using v1/t1=v2/t2
v1=500
v2=?
t1=75=368k
t2=225=498
500/368=v2/498
1.4x498=v2
v2=697.2ml
6 0
3 years ago
A force can exist between two charged particles or objects even if they're as small as subatomic particles. between which of the
murzikaleks [220]
D. a and b because same-charged particles repel each other
8 0
3 years ago
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution
Marianna [84]

Answer:

0.136g

Explanation:

A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?

Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)

Initial mole of Co(NO3)2  =\frac{mass}{molar mass}

=\frac{5.00}{182.94} \\\\=0.02733mol

Mole of Co(NO3)2 in final solution

=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol

Mole of  NO3- in final solution = 2 x Mole of Co(NO3)2

=2\times 0.001093\\\\=0.002186mol

Mass of  NO3- in final solution is mole x Molar mass of NO3

=0.002186\times62.01\\\\=0.136g

6 0
3 years ago
A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o
taurus [48]

Answer:

The concentration the student should write down in her lab is 2.2 mol/L

Explanation:

Atomic mass of the elements are:

Na: 22.989 u

S: 32.065 u

O: 15.999 u

Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.

Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.

For mole(s) of Na2S2O3 = (mass taken)/(molar mass)

= (17.240 g)/(158.105 g/mol) = 0.1090 mole.

Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)

= 0.05029 L.

To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:

= (moles of sodium thiosulfate)/(volume of solution in L)

= (0.1090 mole)/(0.05029 L)

= 2.1674 mol/L

6 0
3 years ago
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