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Dennis_Churaev [7]
3 years ago
14

8.41 x 10 to the -7 grams to microgram

Chemistry
1 answer:
mezya [45]3 years ago
6 0
0.841 micrograms is the answer.

If you have any questions, let me know.
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Helium has a density of 1.79×10^-4 g/mL at standard temperature and pressure. A balloon has a volume of 6.3 liters. Calculate th
djyliett [7]
6,3 L = 6300 mL
..................
1,79×10^-4g -------- 1 mL
Xg --------------------- 6300 mL
X = 0,000179×6300
X = 1,1277g ≈ 1,1g

:•)
7 0
4 years ago
The following equation is balanced correctly. 4C2H6 + 7O2 --&gt; 6H2O + 4CO2<br> True <br> False
Ostrovityanka [42]

Answer: False

Explanation:

4C2H6 + 7O2 --> 6H2O + 4CO2

8 Carbons on the reactant side, but 4 Carbons on the Product

24 Hydrogens on Reactant, 12 H on product

But Oxygen is balanced, 14 on each side

4 0
3 years ago
Read 2 more answers
Based on the Lewis/electron dot representation
blsea [12.9K]

Answer:

A. 1:3

Explanation:

If we look at the ions shown in the image attached to the question, we will notice that we have aluminum (Al^3+), a trivalent ion combining with the iodide ion (I^-).

Aluminum can easily give out its three outermost electrons to three atoms of iodine. If aluminum gives out its three electrons, it achieves the stable octet structure. Iodine atoms have seven electrons in their outermost shell. They only need one more electrons to complete their octet. This one electron can be gotten by the combination of three iodine atoms with one atom of aluminum. One electron each is transferred from the aluminum atom to each iodine atom to form AlI3 with a ratio of 1:3.

7 0
3 years ago
A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
Akimi4 [234]

Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

4 0
4 years ago
How many moles are in 564 grams of Copper
nexus9112 [7]
63.55 is how many miles are in 564 grams of copper
4 0
3 years ago
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