1.

A tall pea plant (Tt) is crossbred with another tall pea plant (Tt). The following Punnett square shows the separated alleles fo

Luba_88 [7]

Answer:

B.

Explanation:

when you use punnett squares, you cross the different genes with themselves. if T is in the first box on both sides, the genome will be TT. if the box lines up to tt, it will be tt. the boxes on the left use the genes' first letter and the boxes on the right use the second letter.

7 0

3 years ago

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50 points I need help on this whole work sheet about converting moles

kozerog [31]

Answer:

Explanation:

11)

Answer:

9.08 mol

Given data:

Number of moles of P₂O₅ = ?

Number of moles of O₂ = 22.7 mol

Solution:

Chemical equation:

4P + 5O₂ → 2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

O₂ : P₂O₅

5 : 2

22.7 : 2/5×22.7 = 9.08

12)

Answer:

7 mol

Given data:

Number of moles of P₂O₅ = ?

Number of moles of P = 14 mol

Solution:

Chemical equation:

4P + 5O₂ → 2P₂O₅

Now we will compare the moles of P₂O₅ with P.

P : P₂O₅

4 : 2

14 : 2/4×14 = 7

13)

Answer:

76.25 mol

Given data:

Number of moles of P = 61 mol

Number of moles of O₂ react = ?

Solution:

Chemical equation:

4P + 5O₂ → 2P₂O₅

Now we will compare the moles of P with O₂.

P : O₂

4 : 5

61 : 5/4×61 = 76.25

14)

Answer:

1.25 mol

Given data:

Number of moles of P₂O₅ = 0.5 mol

Number of moles of O₂ needed = ?

Solution:

Chemical equation:

4P + 5O₂ → 2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

P₂O₅ : O₂

2 : 5

0.5 : 5/2×0.5 = 1.25

15)

Answer:

20 mol

Given data:

Number of moles of P₂O₅ = 8 mol

Number of moles of O₂ needed = ?

Solution:

Chemical equation:

4P + 5O₂ → 2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

P₂O₅ : O₂

2 : 5

8 : 5/2×8 = 20

16)

Answer:

12 mol

Given data:

Number of moles of silver made = ?

Number of moles of Ag₂O = 6 mol

Solution:

Chemical equation:

2Ag₂O → 4Ag + O₂

Now we will compare the moles of Ag with Ag₂O .

Ag₂O : Ag

2 : 4

6 : 4/2×6 = 12

17)

Answer:

25 mol

Given data:

Number of moles of silver made = ?

Number of moles of O₂ produced = 6.25 mol

Solution:

Chemical equation:

2Ag₂O → 4Ag + O₂

Now we will compare the moles of Ag with O₂ .

O₂ : Ag

1 : 4

6.25 : 4×6.25 = 25

18)

Answer:

9.8 mol

Given data:

Number of moles of silver made = ?

Number of moles of O₂ produced = 2.45 mol

Solution:

Chemical equation:

2Ag₂O → 4Ag + O₂

Now we will compare the moles of Ag with O₂ .

O₂ : Ag

1 : 4

2.45 : 4×2.45 = 9.8

19)

Answer:

4.4 mol

Given data:

Number of moles of silver oxide required = ?

Number of moles of O₂ produced = 2.2 mol

Solution:

Chemical equation:

2Ag₂O → 4Ag + O₂

Now we will compare the moles Ag₂O of with O₂ .

O₂ : Ag₂O

1 : 2

2.2 : 2×2.2 = 4.4

20)

Answer:

1.5 mol

Given data:

Number of moles of silver oxide required = ?

Number of moles of O₂ produced = 0.75 mol

Solution:

Chemical equation:

2Ag₂O → 4Ag + O₂

Now we will compare the moles Ag₂O of with O₂ .

O₂ : Ag₂O

1 : 2

0.75 : 2×0.75 = 1.5

8 0

4 years ago

Mention the four characterstics of warm blooded animals.<br>

Igoryamba

Answer:

The four characrteristics of warm blooded animals are÷

Explanation:

i=They can keep its body temperature the same no matter what the outside temperature .

ii=They can maintain a constant body temperature.

iii=They obtain energy from food consumption.

iv=They maintain their body temperature higher than environment.

7 0

3 years ago

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The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

3 years ago

Solve and show work. Li2S + 2 HNO3 --> 2 LiNO3 + H2S (a) Calculate the mass of lithium sulfide that will react with 250 mL of

Elenna [48]

Li2S + 2 HNO3 --> 2 LiNO3 + H2S

Li2 S + H2 N2 O2 --> Li2 N2 O5 + H2 S

Li S + H2 N2 O5 -> Li N2 O5 + H2 S

Li2 S2 + H4 N4 O10 --> Li2 N4 O10 + H4 S2

Li^2 S^2 + H^4 N^4 O^10 --> Li^2 N^4 O^10 + H^4 S^2

7 0

3 years ago