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Alina [70]
3 years ago
6

How are mitosis and meiosis similar? How are they different? Provide 1 way the two processes are the same and 2 ways the 2 proce

sses are different.
Chemistry
1 answer:
swat323 years ago
3 0

Answer:

Similar: both processes of cell division; both processes take place in the nucleus of the cell

Different: mitosis divides into 2, meiosis divides into 4; one is the division of body cells and the other is of specifically sex cells

Explanation:

https://byjus.com/biology/mitosis-and-meiosis/

^ this has more info!

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Many people have a tendency to think that heat and temperature are the same thing. However, this is not true. Heat describes A.
enyata [817]

The correct answer is actually  

B) energy that flows from warmer objects to cooler objects.

because temperature is a measure of the average amount of energy possessed by an object due to the random motions of its particles.  Heat is the energy that flows from warmer objects to cooler objects. Heat cannot flow in the opposite direction.

7 0
3 years ago
Read 2 more answers
Atoms try to have the same number of protrons and electrons. True or false
Dima020 [189]

Answer:

False

Explanation:

8 0
3 years ago
A sample of hydraulic acid has a volume of 2.3L. What is the volume of acid in gallons?
Assoli18 [71]

0.60759572 gal  is the volume of acid in gallons  in a sample of hydraulic acid .

<h3>What do mean by the term "gallons" ?</h3>

The gallon is a unit of measurement for volume and fluid capacity in both the US  units and the British imperial systems of measurement.

In SI base units, 1 Gal is equal to 0.01 m/s²

Since ,

1 L, l = 0.2641720524 gal

now, volume of acid in gallons is given by -

2.3 L, l = 2.3 × 0.2641720524 gal

=0.60759572 gal

Hence , 0.60759572 gal is the volume of acid in gallons  in a sample of hydraulic acid .

Learn more about gallon ,here:

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7 0
1 year ago
Read 2 more answers
In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o
Kitty [74]

Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1}  )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w

The specific gravity of the wood chips is 0.494.

The average volume of a log is

V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2}  \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}

The weight of one log is

M=\rho*V=0.494*\rho_w*12.57  ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm

To provide 3617 ton/day of wood chips, we need

n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min

The feed rate of logs is 14.3 logs/min.

7 0
3 years ago
Read 2 more answers
2C2H6 + 7O2 ------&gt; 4CO2 + 6H2O
Romashka-Z-Leto [24]

We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.

Dividing the mass of each reactant by its molar mass:

(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6

(10 g O2)(31.999 g/mol) = 0.3125 mol O2.

Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.

Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).

So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.

Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.

8 0
3 years ago
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