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2 years ago

What is the mass in grams of 7.5 x 10^15 atoms of nickel?

2 answers:

6 0

Atomic mass Ni = 58.69 a.m.u

58.69 g ----------------- 6.02x10²³ atoms
?? g --------------------- 7.5x10¹⁵ atoms

58.69x (7.5x10¹⁵) / 6.02x10²³

=> 7.31x10⁻⁷ g

vredina [299]2 years ago

5 0

Answer:

The mass of $7,5.10^{15}$ nickel atoms is $7,34.10^{-7}g$ grams

Explanation:

We want to know how many grams $7.5.10^{15}$ nickel atoms equals and we know that one mole of any substance equals 6.021023 atoms

To know how many moles correspond $7.5.10^{15}$ nickel atoms apply a simple rule of three

$6,02.10^{23} atoms\longrightarrow 1 mol\\7,5.10^{15} atoms\longrightarrow x\\x=\frac{(7,5.10^{15})(1mol) }{6,02.10^23atoms} \\x= 1,25.10^{-8} mol$

The constant that relates the mass (grams) and the amount of substance (mol) is the molecular mass of the element The nickel molar mass is 58.69g/mol

to know how many grams of nickel correspond $1,25.10^{-8} mol$ mol we apply a rule of three simple

$1 mol \longrightarrow 58.69 g\\1,25.10^{-8}\longrightarrow x \\x= \frac{(1,25.10^{-8}mol)(58.69g) }{1mol} \\x=7,34.10^{-7}g$

The mass of $7,5.10^{15}$ nickel atoms is $7,34.10^{-7}g$ grams

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Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis

almond37 [142]

Answer: a) $BaCl_2$

b) $BiO_3H_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles$

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = $\frac{0.48}{0.48}=1$

For O = $\frac{0.96}{0.48}=2$

The ratio of Ba: Cl= 1:2

Hence the empirical formula is $BaCl_2$

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= $\frac{0.38}{0.38}=1$

For O = $\frac{1.15}{0.38}=3$

For H= $\frac{1.16}{0.38}=3$

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is $BiO_3H_3$

6 0

2 years ago

Large sharks eat many other marine life animals. The sharks and the animals they eat are all part of which level of organization

Sladkaya [172]

Answer:

the answer is ecosystem i believe.

8 0

2 years ago

Read 2 more answers

For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

SVETLANKA909090 [29]

Answer:

$\frac{1}{24}$

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = $\frac{1}{6}$

So, fraction of the whole athletic field reserved for each fifth class = $\frac{1}{4}(\frac{1}{6})=\frac{1}{24}$

3 0

2 years ago

At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a

djverab [1.8K]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

$Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2$

Where P is the pressure of each compound in equilibrium.

If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:

H2S = 3.00 atm - 2X

H2 = 2X

S2: = X

Replacing:

$3.2x10^{-7} = (2X)^2X / (3-2X)^2$

$3.2x10^{-7} = 4X^3 / 9- 6X+4X^2$

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Solving for X:

X = 0.008945 atm

As in equilibrium, pressure of S2 is X, pressure is 0.008945 atm

7 0

2 years ago

The combustion of 0.374 kg of methane in the presence of excess oxygen produces 0.983 kg of carbon dioxide. What is the percent

anyanavicka [17]

Assuming that the combustion formula is
CH4 + 2O2 --> 2H2O + CO2,

That means for every 1 molecule of methane(CH4) there will be one molecule of carbon dioxide(CO2) produced. Methane molecular weight 16, carbon dioxide molecular weight is 44. Then the percent yield should be:
1 * (0.374/ 16) /(0.983/44)= 0.374*44/ 0.983 * 16= 104.6%

You sure the number is correct? Percent yield should not exceed 100%

5 0

2 years ago