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pochemuha
3 years ago
6

If a horse is at a stop and the. Starts running is the horse accelerating or decelerating?

Chemistry
2 answers:
Feliz [49]3 years ago
5 0
Accelerating, because it’s going from standing still to running, so the speed increases
pshichka [43]3 years ago
3 0
The horse is accelerating.
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Balance the following equations:
lesya [120]
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7 0
3 years ago
What part of the water cycle involves the leaves of plants
zvonat [6]

Answer: The answer to this question is transpiration.

Explanation: I know this answer because i looked it up in a book. The other explanation is I study about this a lot.

5 0
2 years ago
Write the balanced chemical equation for the reaction between aqueous sodium carbonate and
ioda

Answer:

2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)

Explanation:

This question is asking to write and balance an equation between between aqueous sodium carbonate (Na2CO3) and aqueous nitric acid (HNO3). The equation is as follows:

HNO3 (aq) + Na2CO3 (aq) → NaNO3 (aq) + CO2 (g) + H2O (l)

However, this equation is not balanced as the number of atoms of each element must be the same on both sides of the equation. To balance the equation, one will make use of coefficients as follows:

2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)

7 0
3 years ago
When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.
sertanlavr [38]

Answer : The enthalpy change for the solution is 42.8 kJ/mol

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 15.8J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m = mass of water = 100.0 g

\Delta T = change in temperature = T_2-T_1=(32.0-23.9)=8.1^oC

Now put all the given values in the above formula, we get:

q=[(15.8J/^oC\times 8.1^oC)+(100.0g\times 4.18J/g^oC\times 8.1^oC)]

q=3513.8J=3.5138kJ        (1 kJ = 1000 J)

Now we have to calculate the enthalpy change for the solution.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 3.5138 kJ

m = mass of NaOH = 3.25 g

Molar mass of NaOH = 40 g/mole

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{3.25g}{40g/mole}=0.0812mole

Now,

\Delta H=\frac{3.5138kJ}{0.0821mole}=42.8kJ/mol

Therefore, the enthalpy change for the solution is 42.8 kJ/mol

4 0
3 years ago
What Newton’s law is this a example of:
Fantom [35]

Answer:

Newton's third law

Explanation:

every action has an equal and opposite reaction.

8 0
3 years ago
Read 2 more answers
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