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alex41 [277]
3 years ago
8

Which one of the following species is diamagnetic: a. F b. Zn2+ C. Cu2+ d. Ca+​

Chemistry
1 answer:
diamong [38]3 years ago
3 0

Answer:

The correct answer to this question is C

Explanation:

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Imagine that you are outside in the cold. Your nose starts to tingle and hurt. What might you be experiencing?
KATRIN_1 [288]
D. Being cold temperatures can result in a cold nose. With prolonged exposure The body will start to lose heat faster than it can generate it, this is the result of hypothermia.
6 0
3 years ago
Read 2 more answers
Write a balanced equation for the reaction of ca with hcl
Aleksandr-060686 [28]
Ca(s)+2Hcl(aq) ------>CaCl2(s)+H2(g)
3 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
A sulfur ion has 16 protons and 18 electrons. Find the net charge
andrezito [222]

I think its 2-, it gained 2 electrons.

5 0
4 years ago
Can someone show me a step by step with the answer?
Georgia [21]

Answer:

6 mols HCl

Explanation:

(I'm an AP chemistry student but not perfect at this stuff)

you can use the Molarity=Moles/L equation here:

(6M)=(moles HCl)/(1L)

divide by 1 on both sides to isolate the moles of HCl

this gets you 6 moles HCl.

Again I'm just a student so my answer might be wrong, but this question should just require the M=mols/L equation :).

8 0
3 years ago
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