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3 years ago

9

Cobalt has an atomic number of 27 and atomic mass of 59. How many neutrons does cobalt have?

1 answer:

elena-s [515]3 years ago

3 0

Answer:

32

Explanation:

59-27

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100 POINTS BRAINLIEST PICTURE BELOW

Paha777 [63]

Answer:

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Explanation:

Sparking can be caused by defective cooktop elements, loose connectors, a wire with frayed insulation or a broken wire arcing over to the appliance metal frame, says Appliance Repair. If the cause of the sparking is a broken wire or a wire with frayed insulation, do not attempt to splice the wire; you must replace it. Repair Clinic suggests checking the spark ignition switches, as they might be defective and in need of replacing.

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A hot lump of 30.5 g of iron at an initial temperature of 52.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to r

slava [35]

Answer:

26.7°C

Explanation:

Using the formula; Q = m × c × ΔT

Where; Q = amount of heat

m = mass

c = specific heat

ΔT = change in temperature

In this question involving iron placed into water, the Qwater = Qiron

For water; m= 50g, c = 4.18 J/g°C, Initial temp= 25°C, final temp=?

For iron; m = 30.5g, c = 0.449J/g°C, Initial temp= 52.7°C, final temp=?

Qwater = -(Qiron)

m × c × ΔT (water) =- {m × c × ΔT (iron)}

50 × 4.18 × (T - 25) = - {30.5 × 0.449 × (T - 52.7)}

209 (T - 25) = - {13.6945 (T - 52.7)}

209T - 5225 = -13.6945T + 721.7

209T + 13.6945T = 5225 + 721.7

222.6945T = 5946.7

T = 5946.7/222.6945

T = 26.7

Hence, the final temperature of water and iron is 26.7°C

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3 years ago

The ΔHcomb value for anethole is -5539 kJ/mol. Assume 0.840 g of anethole is combusted in a calorimeter whose heat capacity (Cal

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Answer:

Final temperature of calorimeter is 25.36^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.840 g of anethole = $\frac{0.840}{148.2}moles$ of anethole = 0.00567 moles of anethole

1 mol of anethole releases 5539 kJ of heat upon combustion

So, 0.00567 moles of anethole release $(5539\times 0.00567)kJ$ of heat or 31.41 kJ of heat

6.60 kJ of heat increases $1^{0}\textrm{C}$ temperature of calorimeter.

So, 31.41 kJ of heat increases $(\frac{1}{6.60}\times 31.41)^{0}\textrm{C}$ or $4.76^{0}\textrm{C}$ temperature of calorimeter

So, the final temperature of calorimeter = $(20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}$

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