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2 years ago

How many pairs of homologous chromosomes are found in human body cells

Chemistry

1 answer:

suter [353]2 years ago

3 0

There are 22 pairs of <span>homologous chromosomes are found in human body cells.</span>

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When you place a glass over a candle and the flame goes out, the limiting reactant is, oxygen wax energy water

jeka57 [31]

Answer: oxygen

Explanation:

4 0

2 years ago

Bubba falls out of a plane flying over his peanut field. What will bubba speed be after falling for 8 seconds?

chubhunter [2.5K]

Variables we know:

t = 8 seconds

Vi = 0 m/s

g = -9.81

Δy = ?

Vf = ?

Equation we will be using to solve for Vf: Vf = Vi + gt

Steps to solve:

Vf = (0) + (-9.81)(8)

Vf = -78.48 m/s

Hope this helps!! :)

5 0

2 years ago

Read 2 more answers

If the equilibrium constant of the reaction is 0.85, then which statement is true if the mass of A is 10.5 grams; the density of

snow_tiger [21]

Answer:

E. Q < K and reaction shifts right

Explanation:

Step 1: Write the balanced equation

A(s) + 3 B(l) ⇄ 2(aq) + D(aq)

Step 2: Calculate the reaction quotient (Q)

The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.

Q = [C]² × [D]

Q = 0.64² × 0.38

Q = 0.15

Step 3: Compare Q with K and determine in which direction will shift the reaction

Since Q < K, the reaction will shift to the right to attain the equilibrium.

8 0

2 years ago

How many molecules are in 7.34 g of Na2CO3

kenny6666 [7]

Answer:

The answer is 105.98844.

<h3>Explanation: </h3>

We assume you are converting between grams Na2CO3 and mole. You can view more details on each measurement unit: molecular weight of Na2CO3 or mol This compound is also known as Sodium Carbonate.

7 0

2 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

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

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

2 years ago