Answer:
because the acid properties of aspirin may be problematic.
Answer:
1. All red calves i.e. RR
2. All roan calves i.e RW
3. 2 red calves (RR) and two roan calves (RW)
Explanation:
According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):
1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR
2) A red (RR) bullis mated with white (WW) cow: All roan calves i.e RW
3) A roan bull (RW) is mated with red (RR) cow: 2 red calves (RR) and two roan calves (RW).
Answer: The space occupied by the gas at 400 torr and
is 250 mL.
Explanation:
Given:
= 250 mL,
= 800 torr, 
,
= 400 torr, 
Formula used is as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that space occupied by the gas at 400 torr and
is 250 mL.
Answer:
the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C
Explanation:
Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.
The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(
)
Substituting the values into the equation, it becomes
E = σ/ε₀(
) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(
) = 7.12 × 10⁵(
) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C
Therefore, the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C
- C_5H_8+13/2O_2—»5CO_2+4H_2O
Balanced one
- 2C_5H_8+13O_2—»10CO_2+8H_2O
Moles of Pentyne
- Given mass/Molarmass
- 34/68
- 0.5mol
Moles of H_2O
1mol releases 241.8KJ
2mol releases 241.8(2)=483.6KJ