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3 years ago

Sex cells are produced during the process of

Chemistry

2 answers:

Svetradugi [14.3K]3 years ago

6 0

Sex cells (also known as gametes) are produced during the process of meiosis

11Alexandr11 [23.1K]3 years ago

5 0

Mitosis is the process

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The atomic mass for Mg is O 55 O 54 O 60

Ivenika [448]

The answer is 60 (C)

3 0

4 years ago

The city council of a small town wanted to add fluoride to their water so that all the residents would have healthier teeth. The

azamat

Answer:NaF is ionic. NF3 is covalent. SiF4 is ionic. CaF2 is Ionic and NH4F is also ionic. Ionic compounds transfer electrons whereas covalent compounds share electrons hence the word "co". Also, ionic compounds are formed with metal and nonmetal. Where a covalent is with 2 nonmetals. Only ionic compounds would produce fluoride in water because ionic compounds can dissolve in water and covalent compounds cant.

3 0

3 years ago

The moon is 250,000 miles away. How many inches is it from earth?

Lapatulllka [165]

Answer:

1.584e10 = 15,840,000,000,000

Explanation:

250,000 miles

multiply the length value by 63360

25e4 x 63360

= 1.584e10

3 0

3 years ago

Read 2 more answers

Consider the combustion of methanol at some high temperature in a constant-pressure reaction chamber: 2ch3oh (g) + 3o2 (g) \long

Monica [59]

The balanced reaction is

2CH3OH (g) + 3O2 (g) ⟶ 2CO2 (g) + 4H2O (g)

as per equation two moles of methanol gas will react with 3 moles of oxygen

one mole of gas occupies 22.4 L of volume

so the moles and volume goes in same ratio

it means two unit volume of methanol will react with three unit volume of oxygen

therefore 1L of methanol gas will react with 3 /2 L of oxygen

Or 18 L of methanol gas will react with 3 x 18 /2 = 27 L of oxygen

So here oxygen is limiting reagent

As per balanced equation

10L of oxygen will react with = 2 X 10 /3 L of methanol = 6.67 L of methanol gas to give 6.67 L of CO2 gas and 13.33 L of water gas

So overall there will be = 18 - 6.67 L of left out methanol = 11.33 L

And 6.67 L of CO2 + 13.33 L of water = 20 L

Total volume of gas = 11.33+ 20 = 31.33 L

8 0

3 years ago

Read 2 more answers

The product gas is then passed through a concentrated solution of KOH to remove the CO2. After passage through the KOH solution,

Kamila [148]

Answer: The mass percent of nitrogen gas in the compound is 13.3 %

Explanation:

Assuming the chemical equation of the compound forming product gases is:

$\text{Compound}\xrightarrow[CuO(s)]{Hot}N_2(g)+CO_2(g)+H_2O(g)$

Now, the product gases are treated with KOH to remove carbon dioxide.

We are given:

$p_{Total}=726torr\\P_{water}=23.8torr\\$

So, pressure of nitrogen gas will be = $p_{Total}-p_{water}=726-23.8=702.2torr$

To calculate the number of moles of nitrogen, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of nitrogen gas = 702.2 torr = 0.924 atm (Conversion factor: 1 atm = 760 torr)

V = Volume of nitrogen gas = 31.8 mL = 0.0318 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of nitrogen gas = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$0.924atm\times 0.0318L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\n_{mix}=\frac{0.924\times 0.0318}{0.0821\times 298}=0.0012mol$

To calculate the mass of nitrogen gas, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.0012 moles

Putting values in above equation, we get:

$0.0012mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.0012mol\times 28g/mol)=0.0336g$

To calculate the mass percent of nitrogen gas in compound, we use the equation:

$\text{Mass percent of nitrogen gas}=\frac{\text{Mass of nitrogen gas}}{\text{Mass of compound}}\times 100$

Mass of compound = 0.253 g

Mass of nitrogen gas = 0.0336 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{0.0336g}{0.253g}\times 100=13.3\%$

Hence, the mass percent of nitrogen gas in the compound is 13.3 %

8 0

4 years ago