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Ne4ueva [31]
2 years ago
8

How many moles are equal to 5.82 x 10^23 atoms of tungsten (w)

Chemistry
2 answers:
fiasKO [112]2 years ago
6 0

Answer:

0.97 mole

Explanation:

1 mole will give 6.02×10^23 atoms

Xmole of tungsten will give 5.82×10^23 atom of tungsten

X= 5.82×10^23/ 6.02×10^23

X = 0.97 moles of tungsten

nexus9112 [7]2 years ago
4 0

Answer:

0.967 moles

Explanation:

number of moles = number of atoms / avogadro's number

We are given;

Number of atoms = 5.82×10^23

Avogadro's number = 6.02×10^23

Therefore, number of moles

= (5.82×10^23)/(6.02×10^23)

=0.967mol

Therefore, 0.967 moles are equal to 5.82×10^23 atoms

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How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?
maria [59]

Mass of lead (II) chromate is 51 g. The molecular formula is PbCrO_{4} and its molar mass is 323.2 g/mol

Number of moles can be calculated using the following formula:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Putting the values,

n=\frac{(51 g}{323.1937 g/mol}=0.1578 mol

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.

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2 years ago
What element is used in bright flashing advertising signs
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The element used in advertising signs is neon.
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Which species does not have a noble gas electron
lara31 [8.8K]
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A 63.5 g sample of an unidentified metal absorbs 355 ) of heat when its temperature changes
insens350 [35]

0.208 is the specific heat capacity of the metal.

Explanation:

Given:

mass (m)  = 63.5 grams 0R 0.0635 kg

Heat absorbed (q) = 355 Joules

Δ T (change in temperature) = 4.56 degrees or 273.15+4.56 = 268.59 K

cp (specific heat capacity) = ?

the formula used for heat absorbed  and to calculate specific heat capacity of a substance will be calculated by using the equation:

q = mc Δ T

c = \frac{q}{mΔ T}

c = \frac{355}{63.5X 268.59}

 = 0.208 J/gm K

specific heat capacity of 0.208 J/gm K

The specific heat capacity is defined as  the heat required to raise the temperature of a substance which is 1 gram. The temperature is in Kelvin and energy required is in joules.

 

5 0
2 years ago
Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol
prohojiy [21]

Hey there!:

From the given data ;

Reaction  volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar  = g/L * mol/g

[E]t  = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence   kcat of   xyzase is  5.7*10¹⁰ s⁻¹


Hope that helps!



4 0
2 years ago
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