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ch4aika [34]
3 years ago
9

The daily value for phosphorus is a 700 mg how many grams of phosphorus are recommended​

Chemistry
1 answer:
-BARSIC- [3]3 years ago
5 0

The amount of grams of phosphorus that are recommended is 0.700 g

Why?

This is an unit conversion problem. To convert between units we use conversion factors. In this case the conversion factor we want is from mg to g, and this is relatively simple: 1000 mg = 1 g. Now, we arrange the quantities in the following way:

700 mg P*\frac{1g}{1000mg}=0.700 g

Conversion factors are really useful for converting units! Just remember to put the unit you don't want at the bottom, and the one you want at the top.

Have a nice day!

#LearnwithBrainly

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Why are the alkali metals and the halogens very reactive
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Answer:

Answer in explanation

Explanation:

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Alkali metals belong to group 1 of the periodic table while halogens belong to group 17 of the periodic table. This means they are just one electron away from achieving the stability of a noble gas configuration. While alkali metals need to lose one electron to form a univalent positive ion, halogens news to gain one electron to form a univalent negative ion.

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2 years ago
1. El metodo cientifico pretende sistematizar el proceso de obtención del conocimiento cientifico, evitar la subjetividad en los
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Answer:

La afirmación es verdadera.

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2 years ago
Answer the questions in the table below about the shape of the phosphorus hexachloride (PCI^1_6) anion. What word or two-word ph
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This can be seen the image attached.

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6 0
3 years ago
179.1 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 16.1oC. After 306.9 g
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Answer:

the specific heat of the unknown compound is c_u=0.412J/g \cdot C

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                \Delta T = T_2 -T_1

Substituting 16.1°C for T_1 and 27.4°C for T_2

                \Delta T = 27.4 - 16.1

                       =11.3^oC

Generally the change in temperature of  unknown compound is evaluated as

                  \Delta T_u = T_3 -T_2

Substituting 27.4°C for T_2 and 94.3°C for T_3

                                    \Delta T = 94.3 - 27.4

                                           =66.9^oC

Since there is an increase in temperature then heat is gained by water and this can be evaluated as

               H_w = mc_w \Delta T

Substituting 179.1 g  for m , 4.18 J/g.C for c_w(specific heat of water)

             H_w = 4.18 * 179.1 * 11.3

                   = 8459.6J

Since there is a decrease in temperature then heat is lost by unknown compound and this can be evaluated as

                    H_u = m_uc_u \Delta T_u

By conservation of energy law

       Heat lost  = Heat gained  

Substituting 306.9 g  for m_u , 8459.6J for H_u

           8459.6 = 306.9 * c_u * 66.9

  Therefore     c_u = \frac{8459.6}{308.9 *66.9}

                           =0.412J/g \cdot C

                   

4 0
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