One of the results is that the moon is near the earth and the other one, the oceans tide. Even though the earth can hold any object within
ts proximity, the ocean is partly attracted due to its liquid property. At night, the ocean tends to be attracted to the moon by creating a bulge and assigning it as ‘high tide’. This is due to the strong gravitational pull of th moon to the earth.
I hope this helps!
This might be right..
Answer:
2Al+ 6HNO3 ---- 3H2 + 2Al(NO3)3
Explanation:
Put coefficient a,b,c, and d for calculation:
a Al + b HNO3 = c H2 + d Al(NO3)3
for Al: a = d
for H: b = 2c
for N: b = 3d
for O: 3b = 9d
Suppose a=1, then d=1, b=3, c=3/2
multiply 2 to make all natural number, a=2, then b=6, c=3, d=2
A solution is prepared by adding 1.43 mol of potassium chloride (kcl) to 889 g of water. The concentration of kcl is 1.61 molal.
mol of Kcl (potassium chloride)= 1.43
water = 889 g
the formula for calculating molality is:
molality = moles of solute/kilograms of solvent
1kg = 1000g so, 889g = 0.889kg
m = 1.43/0.889 = 1.61 molal
Answer:
To have the electronic configuration equal to 1s²2s²2p⁶3s²3p⁶4s²3d⁷, the chemical element must have an electrical charge equal to 27, that is, it must have 27 electrons, such as Cobalt (Co), for example.
Explanation:
The electronic configuration shown in the question above is known as the Linus Pauling distribution and represents the energy sub-levels that an electrically charged atom can have in relation to the amount of electrons it has.
The layers sub-levels are presented in the following order 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹º 4p⁶ 5s² 4d¹º 5p⁶ 6s² 4f14 5d¹º 6p⁶ 7s² 5f14 6d¹º 7p⁶. Where the small numbers represent the number of electrons in each sub-level and the large numbers represent the layers of electronic distribution.
Accordingly, we can see that an atom that has the configuration 1s²2s²2p⁶3s²3p⁶4s²3d⁷ has 27 electrons, like Cobalt.
Answer:
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
Explanation:
Step 1: Data given
Nitric acid = HNO3
Molar mass of H = 1.01 g/mol
Molar mass of N = 14.0 g/mol
Molar mass O = 16.0 g/mol
Number of moles nitric acid (HNO3) = 0.25 moles
Molairty = 0.10 M
Step 2: Calculate molar mass of nitric acid
Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)
Molar mass HNO3 = 1.01 + 14.0 + 3*16.0
Molar mass HNO3 = 63.01 g/mol
Step 3: Calculate mass of solute use
Mass HNO3 = moles HNO3 * molar mass HNO3
Mass HNO3 = 0.25 moles * 63.01 g/mol
Mass HNO3 = 15.75 grams
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
