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tigry1 [53]
2 years ago
5

Why do you think a virus’s reproduction cycle is called the “lytic cycle”

Chemistry
1 answer:
AURORKA [14]2 years ago
6 0

Answer:

The lytic cycle is named for the process of lysis, which occurs when a virus has infected a cell, replicated new virus particles, and bursts through the cell membrane. This releases the new virions, or virus complexes, so they can infect more cells

Explanation:

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Which of the following is not an example of temperature abuse ?
AnnyKZ [126]
The following of what exactly?
6 0
1 year ago
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You are experimenting on the effect of temperature on the rate of reaction between hydrochloric acid (HCI) and potassium iodide
Paraphin [41]
320,512.8 collisions

Easy trick- 273x = 250,000 find the x value and times it by 350.
4 0
1 year ago
1.00 g of a compound is combusted in oxygen and found to give 3.14g of CO2 and 1.29 g of H2O. From these data we can tell thatA.
lora16 [44]

Answer:

the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula

Explanation:

Mass of CO2 obtained = 3.14 g

Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol

The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g

Mass of H2O obtained = 1.29 g

Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol

The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g

% by mass of carbon = 0.857/1 ×100 = 85.7 %

% by mass of hydrogen = 0.0717/1 × 100 = 7.17%

Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %

Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.

5 0
1 year ago
Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon
Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: Cu_2O\text{ and }CuO

  • <u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles

  • <u>Step 2:</u> 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 2.26 moles.

For Carbon = \frac{2.26}{2.26}=1

For Oxygen  = \frac{4.55}{2.26}=2.01\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is CO_2

  • <u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles

  • <u>Step 2:</u> 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 3.57 moles.

For Carbon = \frac{3.57}{3.57}=1

For Oxygen  = \frac{3.57}{3.57}=1

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is CO

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = \frac{1}{2}\times 1=\frac{1}{2}

Now, taking the ratio of carbon atoms in both the samples, we get:

C_1:C_2=\frac{1}{2}:1=1:2

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0
1 year ago
identify A and B, isomers of molecular formula C3H4Cl2, from the given 1H NMR data: Compound A exhibits peaks at 1.75 (doublet,
siniylev [52]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

3 0
1 year ago
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