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aliya0001 [1]
2 years ago
6

A virus has a mass of ×9.010−12mg and an oil tanker has a mass of ×3.0107kg . Use this information to answer the questions below

. Be sure your answers have the correct number of significant digits.What is the mass of one mole of viruses in grams?
Chemistry
1 answer:
puteri [66]2 years ago
6 0

Answer: Mass of one mole of viruses in grams is 54\times 10^{8}  

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

Given : One virus has mass of = 9.0\times 10^{-12}mg=9.0\times 10^{-15}g     1mg=10^{-3}g

One mole of virus 6.023\times 10^{23} has mass of = \frac{9.0\times 10^{-15}}{1}\times 6.023\times 10^{23}=54\times 10^{8}g  

Thus mass of one mole of viruses in grams is 54\times 10^{8}  

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Give the nuclide symbol for an atom that has<br> mass number 64 and 35 neutrons.
azamat
The mass number represents the summation of the number of protons and the number of neutrons in the nucleus of an atom.
We are given that the mass number is 64 and that the number of neutrons is 35. Therefore:
number of protons = 64 - 35 = 29 protons

In ground state, number of protons in an atom is equal to the number of electrons. Therefore,
number of electrons = 29 electron

Using the periodic table, we will find that the element that has 29 electrons in ground state is copper.
The nuclide symbol of copper is shown in the attached image.

5 0
2 years ago
If 75.0% of the isotopes of an element have a mass of 35.0 amu and 25.0% of the isotopes have a mass of 37.0amu what is the atom
Norma-Jean [14]
 <span>Calculating average atomic mass is exactly like calculating a weighted average. Perform the following calculation: 

(mass1)(percentage1) + (mass2)(percentage2) = average atomic mass 

(35.0)(0.75) + (37.0)(0.25) = average atomic mass 

Make sure your percentages are in decimal form for this calculation. 

One of the other answers given is correct, though the explanation is lacking a bit. Two of the answers can be eliminated immediately: 35.0 amu and 37.0 amu cannot be the average. If the mixture of isotopes was 50% and 50%, then 36.0 amu would be correct; however, the mixture is 75/25. This leaves only one possible answer choice.</span>
8 0
2 years ago
Read 2 more answers
A student requires 2.00 L of 0.100 M NH4NO3 from a 1.75 M NH4NO3 stock solution. What is the correct way to get the solution?
AysviL [449]

Answer:

not sure, but B

Explanation:

7 0
2 years ago
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If you perform the experiment described in investigation #15 by mixing 10 g of glue with 13 g of water and 8 g of sodium borate
Phantasy [73]

10 g of glue with 13 g of water ,

Mass ratio of the material can be calculated as:    

\frac{Mass of glue}{Mass of glue + Mass of water}  

\frac{10 g}{10 g + 13 g }  

\frac{10 g}{23 g }

8 g of sodium borate suspended in 11 g of water, mass ratio can be calculated as:

\frac{Mass of sodium borate}{Mass of sodium borate + Mass of water}  

\frac{ 8 g}{ 8 g + 11 g }  

\frac{8 g}{19 g }

3 0
2 years ago
Suppose you had a balloon containing 1 mole of helium at STP and a balloon containing 1 mole of oxygen at STP. Which statement(s
AleksAgata [21]

Answer:

The true statement  is option A.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas =  1 atm

V = Volume of gas = ?

n = number of moles of gas = 1 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 273.15 K

V=\frac{nRT}{P}=\frac{1 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}

V = 22.42 L

This means that 1 mole of an ideal gas at STP occupies 22.42 liters of volume.

So, 1 mole of helium gas and 1 mole of oxygen gas will have same value of volume in their respective balloons at STP.

7 0
2 years ago
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