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PIT_PIT [208]
2 years ago
7

Please help sue at 11:59 and I need a good grade

Chemistry
2 answers:
ss7ja [257]2 years ago
7 0

Answer:

Option #2 (Green seeds)

Explanation:

Since both the plants are true-breeding, both plants only have the green allele. Therefore, their offspring must also only have the green allele and will therefore have green seeds.

meriva2 years ago
6 0
It’s most likely for the seeds to be green
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Convert 5.28 x 1019 molecules of C6H1206 to grams.
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Answer:

m=0.0158g

Explanation:

Hello there!

In this case, it is possible to comprehend these mass-particles problems by means of the concept of mole, molar mass and the Avogadro's number because one mole of any substance has 6.022x10²³ particles and have a mass equal to the molar mass.

In such a way, for C₆H₁₂O₆, whose molar mass is about 180.16 g/mol, the referred mass would be:

m=5.28x10^{19}molecules*\frac{1mol}{6.022x10^{23}molecules}*\frac{180.16g}{1mol}\\\\m=0.0158g

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3 years ago
Why is it possible for star that is no longer in space to still appear to shine?
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A sample of water is heated from 60.0 °C to 75.0°C by the addition of 140 j of
kupik [55]

Mass of the water : 2.23 g

<h3>Furter explanation</h3>

Heat

Q = m.c.Δt

m= mass, g

c = heat capacity, for water : 4.18 J/g° C.

ΔT = temperature

Q= 140 J

Δt = 75 - 60 = 15

mass of the water :

\tt m=\dfrac{Q}{c.\Delta T}=\dfrac{140}{4.18\times 15}=2.23~g

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Name two elements in the modern periodic table that break Mendeléey's rule that the elements should be arranged in order of rela
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3 years ago
For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.
Rudik [331]

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

8 0
3 years ago
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