Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Its A. because it measures the rate of the decay of the isotope
Can you tell us what you mean and post the questions ?
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)
D, obviously. All living things are made of one or more cells, therefore non-living things aren’t made of cells
