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2 years ago

What is 0.05 kg mass in milligrams?

Chemistry

2 answers:

Crank2 years ago

7 0

0.05 kg = 50000 milligrams

liberstina [14]2 years ago

3 0

Convert 0.05 Kilograms to Milligrams (kg to mg) with our conversion calculator and conversion tables. To convert 0.05 kg to mg use direct conversion formula below.
0.05 kg = 50000 mg.
You also can convert 0.05 Kilograms to other Weight (popular) units.

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2 years ago

PLEASE HELP ME what would be the mass of 9.03 x 1021 molecules of hydrobromic acid (HBr)?

Nikitich [7]

We know 1 mole of any atom or molecules contains $6.033 \times 10^{23}$ atom or molecules.

1 mole of HBr i.e 81 gm/mol contains $6.033 \times 10^{23}$ atom or molecules.

So, mass of $9.03\times 10^{21}$ molecules is :

$m=\dfrac{81\times 9.03\times 10^{21} }{6.033 \times 10^{23}}\\\\m= 1.21\ gm$

Therefore, mass of $9.03\times 10^{21}$ molecules is 1.21 gm .

Hence, this is the required solution.

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2 years ago

How many moles of Cl- ions are needed to completely combine with 0.25 moles of Mg+2 ions?

Mnenie [13.5K]

The number of moles of Cl- ions needed to combine completely with 0.25 mole of Mg+2 ions is:
0.50.

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2 years ago

What is a small description of a atom

steposvetlana [31]

Answer:

An atom is a particle of matter that uniquely defines achemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons. ... The nucleus is positively charged, and contains one or more relatively heavy particles known as protons and neutrons.

4 0

2 years ago

Read 2 more answers

Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

$Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%$

Best regards!

8 0

2 years ago