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2 years ago

Distinguish air mass and a front

Chemistry

1 answer:

Readme [11.4K]2 years ago

3 0

Air mass is how much matter is in air. Front is the side that something faces forward with.

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36. What is the formula of pentaphosphorus decoxide?<br> A)P5010<br> B)PO 10<br> C)PO<br> D)P10 05

rosijanka [135]

Answer:

A)P5010

Explanation:

Penta- means 5 and deca/deco- means 10

3 0

3 years ago

What type of cryptography uses two keys instead of just one, generating both a private and a public key?

Alexxx [7]

Answer:

It is known as asymmetric key cryptography it is also called public key cryptography.

Explanation:

Asymmetric key cryptography method makes use of two keys.One is used for encryption and the second one for decryption. The public key serves to encrypt plain text or verify a digital signature, while the private key is used to decrypt or decipher the encrypted text or to create a digital signature.

3 0

2 years ago

Convert 6.42 moles of H2O to molecules

Ivahew [28]

Answer:

3.87 x 10^24

Explanation:

Simply multiply the moles by avogadros number

6.42 moles of H2O x 6.022 x 10^23 molecules/1 mole of H2O = 3.87 x 10^24 molecules of H2O

4 0

3 years ago

A hydrogen filled balloon has a volume of 8.3 L at 36°C and 751 mmHg. How many moles of hydrogen are inside the balloon?

tamaranim1 [39]

Answer: I think the formula is PV=nRT and I divide both sides by RT, but this is as far as I can get in my equation before I get stumped: (751 mm Hg) (8.3 L)/ (309 K) Can you help?

Explanation:

8 0

2 years ago

A sample of 53.0 of carbon dioxide was obtained by heating 1.31 g of calcium carbont. What is the percent yield for this reactio

MaRussiya [10]

Answer:

92.04%

Explanation:

Given:

Mass of CO₂ obtained = 53.0 grams

Mass of calcium carbonate heated = 1.31 grams

Now,

the molar mass of the calcium carbonate = 100.08 grams

The number of moles heated in the problem = Mass / Molar mass

= (1.31 grams) / (100.08 grams/moles)

= 0.013088 moles

now,

1 mol of calcium carbonate yields 1 mol of CO₂

thus,

0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂

now,

Theoretical mass of 0.013088 moles of CO₂ will be

= Number of moles × Molar mass of CO₂

= 0.013088 × 44 = 0.5758 grams

Thus, the percent yield for this reaction = $\frac{\textup{Actual yield}}{\textup{Theoretical yield}}\times100$

the percent yield for this reaction = $\frac{0.53}{0.5758}\times100$

the percent yield for this reaction = 92.04%

6 0

2 years ago