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3 years ago

WHICH OF THE GENOTYPES IN NUMBER 1 WOULD BE CONSIDERED PUREBRED

1 answer:

3 0

Purebred — Also called HOMOZYGOUS and consists of gene pairs with genes that are the SAME. Hybrid - Also called HETEROZYGOUS and consists of gene pairs that are D i'FFEREN'i". Genotype is the actual GENE makeup represented by LE'H'ERS.

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A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0

miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

$Q_{metal} - Q_{water} = 0$

$Q_{metal} = Q_{water}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{metal}$ - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

$m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})$

The specific heat of the metal is cleared within equation above:

$c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}$

If we know that $m_{water} = 0.237\,kg$ , $m_{metal} = 0.244\,kg$ , $c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{w,o} = 10\,^{\circ}C$ , $T_{m,o} = 100.10\,^{\circ}C$ and $T = 15.30\,^{\circ}C$ , the specific heat of molybdenum is:

$c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}$

$c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}$

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

5 0

3 years ago

Help on question 1b please? (image attached)

aliya0001 [1]

I believe your answer Is C. An ammonia molecule has a trigonometrical pyramidal shape. Figure C has a has a trigonometrical pyramidal shape.

I hope I help

3 0

3 years ago

The molar solubility of ag2s is 1.26 × 10-16 m in pure water. calculate the ksp for ag2s.

MAXImum [283]

Answer is: Ksp for silver sulfide is 8.00·10⁻⁴⁸.
Reaction of dissociation: Ag₂S(s) → 2Ag⁺(aq) + S²⁻(aq).
s(Ag₂S) = s(S²⁻) = 1.26·10⁻¹⁶ M.
s(Ag⁺) = 2s(Ag₂S) = 2.52·10⁻¹⁶ M; equilibrium concentration of silver cations.
Ksp = s(Ag⁺)² · s(S²⁻).
Ksp = (2.52·10⁻¹⁶ M)² · 1.26·10⁻¹⁶ M.
Ksp = 6.35·10⁻³² M² · 1.26·10⁻¹⁶ M.
Ksp = 8.00·10⁻⁴⁸ M³.

6 0

3 years ago

Which is the largest group of mollusks?

KonstantinChe [14]

Answer:

Gastropoda.

Explanation:

3 0

3 years ago

Read 2 more answers

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a

svetoff [14.1K]

Answer:

The answer to your question is P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

= 373°K

Formula

$(P + \frac{a}{v^{2}} )(v - b) = RT$

Substitution

$(P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)$

Simplify

(P + 0.0094)(22.3829) = 30.586

Solve for P

P + 0.0094 = $\frac{30.586}{22.3829}$

P + 0.0094 = 1.366

P = 1.336 - 0.0094

P = 1.357 atm

7 0

3 years ago