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Masja [62]
3 years ago
13

A body that weighs 6 Newton on the earth will weighs – on the moon in newton​

Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
6 0
An object on the moon is 1/6 of the weight of an object on Earth

To find the answer to your question, divide the number of N on Earth by 6, 6N / 6 = 1N

Hope this helps!
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Which of the following should
Sergeu [11.5K]

Answer:

warm water holds the least amount of dissolved oxygen, so I would assume the answer would be D. a small pond could heat up easily. in addition, the water is calm and not moving in a pond

4 0
3 years ago
Which of these is a compound?<br><br> A. Steel<br><br> B.Sugar<br><br> C.Air<br><br> D.Nitrogen
hjlf

Sugar is a compound made of carbon, hydrogen and oxygen.

8 0
3 years ago
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What characterizes an Arrhenius base?
dybincka [34]

Answer:

An Arrhenius base is a substance that dissociates in water to form hydroxide (OH–) ions. In other words, a base increases the concentration of OH– ions in an aqueous solution.

7 0
2 years ago
Calculate the DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide. DH°f means delta or change
USPshnik [31]

Answer: +178.3 kJ

Explanation:

The chemical equation  follows:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]

We are given:

\Delta H^o_f_{(CaO(s))}=-635.1kJ/mol\\\Delta H^o_f_{(CaCO_3(s))}=-1206.9kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]

The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ

5 0
3 years ago
Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a
GrogVix [38]

Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

8 0
3 years ago
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