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3 years ago

Gabriel and his friend found a piece of metal which they want to

Chemistry

2 answers:

Dominik [7]3 years ago

6 0

50/5.2 that’s the equation that you have to solve then whatever comes out yo calculator is the answer

Crazy boy [7]3 years ago

4 0

Answer:

Zinc is the metal that your looking for there bud

Explanation:

You might be interested in

Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

$Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%$

Best regards!

8 0

3 years ago

Suppose you have 81.8 g of Cd. How many Cd atoms are present? Report your answer in scientific notation using the format of 6.02

Lostsunrise [7]

Answer:

1. number of Cd atoms = 4.384 × 10²³

2. number of moles of Ti = 0.817 mol

3. molar mass of C₇H₆O₃ = 138.077 g/mol

4. moles of Sildenafil = 0.249 mmol

Explanation:

1. given data

mass of Cd = 81.8 g

number of Cd atoms = ?

molar mass of Cd = 112.411 g/mol

Solution

1st we find out the number of moles of Cd

number of moles = mass / molar mass

number of moles of Cd = 81.8 g / 112.411 g.mol⁻

number of moles of Cd = 0.728 mol

Now we find out the number of Cd atoms

number of atoms = moles × 6.022 × 10²³

number of Cd atoms = 0.728 (6.022 × 10²³)

number of Cd atoms = 4.384 × 10²³

2. Given data

Mass of Ti = 39.1 g

moles = ?

molar mass Ti = 47.867 g/mol

Solution

Now we will find out the number of moles

number of moles = mass / molar mass

number of moles of Ti = 39.1 g / 47.867 g.mol⁻

number of moles of Ti = 0.817 mol

3. Given data

chemical formula of Salicylic acid = C₇H₆O₃

molar mass of salicylic acid = ?

molar mass of carbon = 12.011

molar mass of hydrogen = 1.008

molar mass of oxygen = 15.999

molar mass of molecule = number of moles of atoms × molar mass of atoms

molar mass of C₇H₆O₃ = 7 (12.011) + 6 (1) + 3 (16)

molar mass of C₇H₆O₃ = 84.077 g + 6 g + 48 g

molar mass of C₇H₆O₃ = 138.077 g/mol

4. Given data

molar mass of Sildenafil = 474.5828 g/mol

mass given = 0.118 g

moles of Sildenafil = ?

Solution

Now we find out the number of moles

number of moles = mass / molar mass

number of moles Sildenafil = 0.118 g / 474.5828 g/mol

number of moles Sildenafil = 0.0003 mol

as this is very small amount we convert it into millimoles

millimoles = moles × 1000

millimoles of Sildenafil = 0.000249 × 1000

millimoles of Sildenafil = 0.249 mmol

4 0

3 years ago

Its atom contains 50 electrons and 70 neutrons

Tcecarenko [31]

Tin-120 because an atom has to have a neutral charge. If an atom has 50 electrons it must have 50 protons unless it's an ion. Add 70 to the atomic mass because it has 70 neutrons on top of 50 protons. That equals to 120. Tin has an atomic number of 50 (meaning it has 50 protons) and it has roughly an atomic mass of about 118. If the given has 70 neutrons then it's a isotope of tin.

4 0

3 years ago

What mass of water would be produced if 1.5 g of ethanoic acid was reacted completely with sodium hydroxide (3 marks)

Zigmanuir [339]

Mass of water : 0.45 g

<h3>Further explanation</h3>

Given

1.5 g ethanoic acid

Required

Mass of water

Solution

Reaction

CH₃COOH + NaOH ⇒H₂O + CH₃COONa

mol CH₃COOH(MW=60 g/mol) :

= mass : MW

= 1.5 : 60

= 0.025

mol H₂O from the equation :

= 0.025(mol ratio CH₃COOH : H₂O= 1 : 1)

Mass of water :

= mol x MW water

= 0.025 x 18 g/mol

= 0.45 g

5 0

3 years ago

What mass, in grams, of NaNo, is needed to make 6.24 Liters of a 0.500 M solution?

mafiozo [28]

Answer:

that hared ldkdjjjdjjdjdjj

3 0

3 years ago