Answer:
The temperature would be 194, 8 K.
Explanation:
We use the formula:
PV=nRT --> T=PV/nR
T= 0,500 atm x 28, 6 L/ 0,895 mol x 0,082 l atm /K mol
<em>T= 194,8494345 K</em>
I could only find 7!
- independent variable
- dependent variable
- control group
- experimental group
- constant
- observation
- inference
Answer:
6.78 × 10⁵ J
Explanation:
Step 1: Given and required data
- Latent heat of vaporization of water (ΔH°vap): 2260 J/g
Step 2: Calculate the heat (Q) required to vaporize 300 grams of water
We will use the following expression.
Q = ΔH°vap × m
Q = 2260 J/g × 300 g
Q = 6.78 × 10⁵ J
6.78 × 10⁵ Joule will be required to convert 300 g of water to steam at 100 °C.
KE=6,250Joules
You plug in your given numbers and remember to square 5 because that’s the speed
Answer:
yes
Explanation:
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