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salantis [7]
3 years ago
7

Can someone please help me with this question? Thanks in advance.

Chemistry
2 answers:
goblinko [34]3 years ago
8 0
Fe2+H4+sulphur2+O8 think of balancing equations is the sum of the equal product of the metal and H,O
sveta [45]3 years ago
6 0

Answer:

1. Fe + H2SO4 = FeSO4 + H2

iron + sulfuric acid = duretter + hydrogen

2. Mg + 2 HNO3 = Mg(NO3)2 + H2

Magnesium + nitric acid = magnesium nitrate +hydrogen

3. Zn + 2 HCl = ZnCl2 + H2

Zinc + hydrogen chloride= zinc chloride+ hydrogen

You might be interested in
If a plant uses 44 grams of carbon dioxide and 18 grams of water to produce 32 grams of oxygen, how many grams of glucose are pr
Tems11 [23]
The balanced equation for the reaction is as follows 
6H₂O + 6CO₂ ---> C₆H₁₂O₆ + 6O₂

number of moles of CO₂ used - 44 g / 44 g/mol = 1 mol
number of moles of H₂O used - 18 g / 18 g/mol = 1 mol
stoichiometry of CO₂ to H₂O is 6:6 = 1:1
1 mol of CO₂ present and 1 mol of H₂O present therefore they are both fully used up in the reaction of molar ratio 1:1
number of moles of O₂ formed - 32 g/ 32 g/mol = 1 mol

stoichiometry of CO₂:H₂O:C₆H₁₂O₆:O₂ is 6:6:1:6
1 mol of CO₂ has reacted with 1 mol of H₂O to form 1 mol of O₂ and x mol of C₆H₁₂O₆

the number of C₆H₁₂O₆ moles is 1/6th of CO₂ moles used up
since CO₂ moles - 1 mol 
therefore C₆H₁₂O₆ moles formed - 1/6 mol = 0.167 mol
mass of glucose formed - 0.167 g x 180 g/mol = 30 g 
therefore 30 g of glucose is formed 
5 0
3 years ago
Review the table detailing the mass of Jupiter’s moons and their distance from the planet. Which of Jupiter’s moons has the grea
olga nikolaevna [1]

The correct answer is Lo.  

The gravitational force between the two components is directly equivalent to the product of their masses and is inversely proportional to the distance separated between them. The largest planet in the solar system is Jupiter. It comprises 75 moons, and out of these moons, the four Galilean moons are very big in mass. These are Lo, Europa, Ganymede, and Callisto.  

Of these Galilean moons, the Lo moon is very close to Jupiter. The Ganymede moon is the largest of all the Galilean moons, but it is situated very far from Jupiter in comparison to Lo. Thus, the force of attraction between the Lo and Jupiter is very high, it exhibits the greatest gravitational force with Jupiter.  

8 0
3 years ago
The pressure in an automobile tire filled with air is 245.0 kPa. If Po2 = 51.3 kPa, Pco2 = 0.10 kPa, and P-others = 2.3 kPa, wha
Kay [80]

Answer:

PN₂ = 191.3 Kpa

Explanation:

Given data:

Total pressure of tire = 245.0 Kpa

Partial pressure of PO₂ = 51.3 Kpa

Partial pressure of PCO₂  = 0.10 Kpa

Partial pressure of others =  2.3 Kpa

Partial pressure of PN₂ = ?

Solution:

According to Dalton law of partial pressure,

The total pressure inside container is equal to the sum of partial pressures of individual gases present in container.

Mathematical expression:

P(total) = P₁ + P₂ + P₃+ ............+Pₙ

Now we will solve this  problem by using this law.

P(total) = PO₂ + PCO₂ + P(others)+ PN₂

245 Kpa = 51.3 Kpa + 0.10 Kpa + 2.3 Kpa + PN₂

245 Kpa = 53.7 Kpa+ PN₂

PN₂ = 245 Kpa -  53.7 Kpa

PN₂ = 191.3 Kpa

6 0
4 years ago
What mass of Na2CO3 is required to create 750 mL of an aqueous solution where the concentration of the carbonate (CO3-2) ions is
Julli [10]

Answer:

139.11 grams

Explanation:

The molarity (M) is equal to m (moles) /L (liters).

Notice how it is in mL so converting it would be 0.75 L (750*10-3)

Now plug in the information

1.75=m/0.75

m=1.3125

Now m is in moles so convert to grams to get the mass

1.3125 mol Na2CO3 * 105.9888 g Na2CO3 = 139.1103

8 0
3 years ago
How many lithium ions are present in 30.0 ml of 0.600 m li2co3 solution?
Nikolay [14]
First; use the relationship of molarity and moles/liter to find the moles of the solution.
so 0.600M ×0.030L = 0.018 moles
Then use the mole to mole ratio of lithium to lithium carbonate
0.18 × (2 Li ÷1 Li2CO3) = 0.036
and then multiply by Avogadro's number to find the ions of lithium
0.036 moles × (6.022×10^{23} ) = 2.167 ×10∧22 ions Li
4 0
3 years ago
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