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Mrac [35]
2 years ago
6

a candle is placed between a concave mirror and its focal point which two of the following best describe the image​

Chemistry
1 answer:
Stels [109]2 years ago
4 0

Answer:

Beyond the centre of curvature

Explanation:

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Questions for buffers lab, please help.
TEA [102]

Answer:

because the acid properties of aspirin may be problematic.

4 0
2 years ago
Determine the possible traits of the calves if:
White raven [17]

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).

7 0
3 years ago
4
STALIN [3.7K]

Answer: The space occupied by the gas at 400 torr and 25^{o}C is 250 mL.

Explanation:

Given: V_{1} = 250 mL,    P_{1} = 800 torr,     T_{1} = 50^{o}C

V_{2} = ?,         P_{2} = 400 torr,        T_{2} = 25^{o}C

Formula used is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{800 torr \times 250 mL}{50^{o}C} = \frac{400 torr \times V_{2}}{25^{o}C}\\V_{2} = 250 mL

Thus, we can conclude that space occupied by the gas at 400 torr and 25^{o}C is 250 mL.

7 0
3 years ago
A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field
maksim [4K]

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(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 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

7 0
3 years ago
Balance the following reaction:
kvv77 [185]
  • 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

  • 8/2(0.5)
  • 4(0.5)
  • 2mol

1mol releases 241.8KJ

2mol releases 241.8(2)=483.6KJ

8 0
2 years ago
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