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3 years ago

Can some one tell me the best position in sex

Chemistry

1 answer:

masya89 [10]3 years ago

8 0

Answer:

speedbump and magician

Explanation:

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What is the concentration of magnesium bromide, in ppm, if 133.4 g MgBr2 dissolved in 1.84 L water. Then solve for the bromine c

Serhud [2]

Answer: 0.0725ppm

Explanation:

133.4g of MgBr2 dissolves in 1.84L of water.

Therefore Xg of MgBr2 will dissolve in 1L of water. i.e

Xg of MgBr2 = 133.4/1.84 = 72.5g

The concentration of MgBr2 is 72.5g/L = 0.0725mg/L

Recall,

1mg/L = 1ppm

Therefore, 0.0725mg/L = 0.0725ppm

7 0

3 years ago

the element silver has an atomic mass of 108 u and consists of two stable isotopes silver-107 and silver-109.

DochEvi [55]

Silver-109 has a mass of 109 amu. It is one of the isotopes of the silver atm.

<h3>What are isotopes?</h3>

The term isotopes has to do with the atoms of the same element that has the same atomic number but different mass numbers. This implies that the isotopes that we have do have the same number of protons but they do not have the same number of neutrons and the fact that they differ in the number of neutrons implies that would have the same chemical property.

We can see from the question that there are two isotopes of silver and tehse isotopes of silver have been identified as silver-107 and silver-109. The isotopes do have the same mass and not the same number of protons.

The percentage abundance of the two isotopes are not the same which means that one is more than the other. We can now see that the mass of the silver 109 in amu is 109 amu.

Learn more about atomic mass:brainly.com/question/5661976

#SPJ1

7 0

2 years ago

Calculate ΔHo for the following reaction ussing the given bond dissociation energiesCH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)BOND

Mazyrski [523]

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

$CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,$ ΔH° = ?

We are given with:

$\Delta H_{O-O}=142 kJ/mol$

$\Delta H_{O=O}=498 kJ/mol$

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{C-H}=411 kJ/mol$

$\Delta H_{C-O}=358 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH° =

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})$