Answer:
As the garter snake can be found almost in any kind of habitat, what makes them be able to survive in any environment include:
1. They hibernate to increase their chances of survival in unfavorable weather conditions.
2. They can blend with the background of any environment especially grass to escape being eaten.
3. They produce an odor that is usually unpleasant especially when about to be attacked.
Explanation:
The garter snakes are distinguished by the three stripes running the length of their body and can often be found in forests, places that are even close to water bodies, and almost any place, even in holes.
Answer:
Explanation:
q = (mass) (temp change) (specific heat)
q = (10000 g) (40 °C) (0.385 J/g⋅°C) = 154000 J = 154 kJ
154 kJ / 2220 kJ/mol = 0.069369369 mol
0.069369369 mol times 44.0962 g/mol = 3.06 g (to three sig figs)
answer choice 4
<h3>The enthalpy of combustion per mole of anthracene : 7064 kj/mol(- sign=exothermic)</h3><h3>Further explanation </h3>
The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released
Q in = Q out
Heat can be calculated using the formula:
Q = mc∆T
Heat released by anthracene= Heat absorbed by water
Heat absorbed by water =
mol of anthracene (MW=178,23 g/mol)
The enthalpy of combustion per mole of anthracene :
Answer:
64.0
Explanation:
2Mg+O2 ---> 2MgO
use dimentional analysis to find the amount of moles of O2 needed first
4.00molMg x 1.00mol O2/ 2.00 mol Mg=. 2.00 mol O2
using the coefficients you can see the mole ratio for O2:Mg the mole ratio is 1:2 which is why there is 1 mole on the top for 2 moles on the bottom. The Mg would cancel and multiply 4 by 1 then divide by 2, or multipy 4 by 1/2
Now that you have the moles of O2 you use the molar mass to find the grams in 2 moles of O2
2.00 mol O2 x 32.0g/1.00 mol = 64.0 g
multiply 2 by 32
Answer:
Explanation:
The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is:
CaCO₃(s) ⟶ CaO(s) + CO₂(g)
ΔH°f/kJ·mol⁻¹: -1207.1 -157.3 -393.5
![\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [-157.3 + (-393.5)] - (-1207.1)\\& = & -550.8 +1207.1\\& = & \textbf{656.3 kJ/mol}\\\end{array}\\\\\text{The enthalpy of decomposition is } \boxed{\textbf{656.3 kJ/mol}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CDelta_%7B%5Ctext%7Br%7D%7DH%5E%7B%5Ccirc%7D%20%26%20%3D%20%26%20%5B-157.3%20%2B%20%28-393.5%29%5D%20-%20%28-1207.1%29%5C%5C%26%20%3D%20%26%20-550.8%20%2B1207.1%5C%5C%26%20%3D%20%26%20%5Ctextbf%7B656.3%20kJ%2Fmol%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5C%5C%5Ctext%7BThe%20enthalpy%20of%20decomposition%20is%20%7D%20%5Cboxed%7B%5Ctextbf%7B656.3%20kJ%2Fmol%7D%7D)
