The answer should be <span>balance electrically
</span><span>Chemical reactions that form ions should have a balanced charge. The example of the reaction is HCl. When forming ions, the equation should be:
HCl => </span>
+
In this case, the hydrogen has one plus charge and chlorine has one negative charge. The resultant should be zero, so it's balanced.
Answer:
B
Explanation:
Soil structure affects water and air movement in a soil, nutrient availability for plants, root growth, and microorganism activity. ... This allows for greater air and water movement and better root growth.
Answer:
0.645 L
Explanation:
To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.
(Step 1)
Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol
Molar Mass (KOH): 56.104 g/mol
19.9 grams KOH 1 mole
-------------------------- x ----------------------- = 0.355 moles KOH
56.014 grams
(Step 2)
Molarity = moles / volume <----- Molarity ratio
0.550 M = 0.355 moles / volume <----- Insert values
(0.550 M) x volume = 0.355 moles <----- Multiply both sides by volume
volume = 0.645 L <----- Divide both sides by 0.550
Hello!
To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.
In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.
Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.
35.0° C - 25.0° C = 10.0° C
We are also given the specific heat and the grams of water. With that, we can substitute the given values into the equation and multiply.
q = 125 g × 4.184 J/g °C × 10.0° C
q = 523 J/°C × 10.0° C
q = 5230 J
Therefore, it will take 5230 joules (J) to raise the temperature of the water.
Given that
Mass of water = 65.34 g
Amount of heat = mass of water * specific heat (temperature change
)
= 65.34 g * 4.184 J / g-C ( 21.75-18.43 )C
= 907.63 J
= 0.908 KJ
And
1 cal = 4.186798 J
907.63 J * 1 cal / 4.186798 J =216.78 cal
Or0.218 kcal
