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elixir [45]
3 years ago
9

If you have 45.6mL of a liquid and its

Chemistry
1 answer:
tankabanditka [31]3 years ago
4 0
The density of the liquid is 1.027g/cm³

D=M/V
D=46.83/45.6
D=1.027g/cm³
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How many moles are in 6.777 x 1023 atoms of CO2. Watch your significant figures. Question 2 options: 1 atoms CO2 1.125 moles CO2
Archy [21]

Answer:

1.13moles

Explanation:

Given parameters:

Number of atoms  = 6.777 x 10²³ atoms

Unknown:

Number of moles  = ?

Solution:

A mole of a substance contains the avogadro's number of particles

       6.02  x  10²³ particles   = 1 mole

      6.777 x 10²³ atoms will contain \frac{ 6.777 x 10^{23}  }{6.02 x 10^{23} }   = 1.13moles

6 0
3 years ago
Calculation: If a compound is 14.40% hydrogen by mass and 85.60% carbon by mass, what is the empirical formula?
IgorLugansk [536]

Answer:

CH2

Explanation:

We are given that a compound is 14.40% hydrogen and 85.60% carbon.

Let the mass of the substance be 100g.

Mass of the hydrogen: 14.40% of 100 = 14.40 g

Mass of the carbon: 85.60% of 100 = 85.60 g

Now, let's find the moles of hydrogen:

14.40 g*\frac{1 mole}{1gH} = 14.40 mole

Moles of carbon:

85.60g*\frac{1 mol}{12gC} = 7.13mol

Let's put these in a ratio and simplify:

7.13 mole C: 14.40 mole H

1 mole C: 2 mole H

Therefore, the empirical formula of this compound is CH2.

Hope this helps!! If you have any questions about my work, please let me know in the comments!

3 0
2 years ago
How many molecules are there in 2.30 grams of NH3
mixer [17]
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We calculate as follows:

2.30 g NH3 (1 mol / 17.03 g ) (</span>6.022 x 10^23 molecules / 1 mol ) = 8.13x10^22 molecules
7 0
3 years ago
What type of galaxy do we live in
strojnjashka [21]
We live in the Milky Way Galaxy
7 0
3 years ago
PLEASE SOMEONE HELP ME WITH THIS!
Bad White [126]

Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877  and 4) 1.04x10^{3\\}

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x  \frac{1 L}{1000 mL} = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x \frac{10 dg}{ 1 g} = 1.291 dg

For the next exercises, you need to follow some rules:

1.  All numbers  that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062  - 638.1848  = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.  

1.04x10^{3}

4 0
3 years ago
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