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How many significant figures are in 5.76 e 15

Dvinal [7]

Answer: 3 significant figures

Explanation: because in scientific notation only the before and after decimal number are considered but in standard form all numbers including exponents are all significant figures

5 0

3 years ago

Read 2 more answers

2. Calculate the mass of 3.47x1023 gold atoms.

lapo4ka [179]

3.47 x $10^{23}$ atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x $10^{23}$

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X $10^{23}$

from the relation,

1 mole of element contains 6.022 x $10^{23}$ atoms.

so no of moles of gold given = $\frac{3.47 X 10^{23} }{6.022 X 10^{23} }$

0.57 moles of gold.

from the relation:

number of moles = $\frac{mass}{atomic mass of 1 mole}$

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x $10^{23}$ atoms of gold have mass of 113.44 grams

3 0

3 years ago

In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so

gulaghasi [49]

<u>Answer:</u> The moles of hydroxide ions present in the sample is 0.0008 moles

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL$

Putting values in above equation, we get:

$1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M$

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

$0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol$

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0

3 years ago

Which of these are examples of passive transport? A. osmosis B.diffusion c. endocytosis d.8 mitosis

AfilCa [17]

Answer:

Passive Transport

Explanation:

The three examples of passive transport are

Diffuison

Osmosis

facilated diffuison

So the answer can be A or B

4 0

2 years ago

What is the relationship between atoms and mass?

castortr0y [4]

Answer:

Explanation:

A property closely related to an atom's mass number is its atomic mass. The atomic mass of a single atom is simply its total mass and is typically expressed in atomic mass units or amu.

6 0

3 years ago