a= 60
b=60
c= 40
d=20
if this is what u mean then i think its correct :)
p.s a parallelogram adds up to 180
Answer:
d=2760,9m
Explanation:
<em>The gravitational force between two equal masses is defined as </em>
<em>F=G.m^2/d^2 where m is mass, G is the universal gravitation ( approximately= 6,67x10^-11N.m^2/kg^2) constant , F is the attractive (or repulsive) force and d is the distance between the masses. </em>
<em>Substitude the values of the quantities into the equation </em>
<em>d^2= G.m^2/F</em>
<em>d= 2760,9m</em>
<em />
<h3>
Answer:</h3>
8.09 × 10²² formula units
<h3>
Explanation:</h3>
Mass of MgCl₂ is 12.8 g
Required to calculate the formula units
1 mole of a compound contains the Avogadro's number formula units
That is;
1 mole = 6.022 × 10²³ formula units
<h3>
Step 1: Calculate the moles of MgCl₂ in 12.8 g </h3>
The molar mass of MgCl₂ is 95.211 g/mol
Number of moles = Mass ÷ molar mass
Moles of MgCl₂ = 12.8 g ÷ 95.211 g/mol
= 0.1344 mole
<h3>Step 2: Calculate the formula units of MgCl₂</h3>
1 mole = 6.022 × 10²³ formula units
Therefore;
Number of formula units = Moles × 6.022 × 10²³ formula units
= 0.1344 mole × 6.022 × 10²³ formula units
= 8.09 × 10²² formula units
Therefore, there are 8.09 × 10²² formula units in 12.8 g of MgCl₂
Answer:
250 °C.
Explanation:
From the question given above, the following data were:
Heat (Q) required = 20.0 kilocal
Mass (M) of water = 80 g
Change in temperature (ΔT) =.?
Next, we shall convert 20.0 Kilocalories to calories. This can be obtained as shown below:
1 kilocal = 1000 Cal
Therefore,
20 kilocal = 20 kilocal × 1000 Cal / 1 kilocal
20 kilocal = 20000 calories
Therefore, 20 kilocalorie is equivalent to 20000 calories.
Finally, we shall determine the change in temperature observed as follow:
Heat (Q) required = 20000 Cal
Mass (M) of water = 80 g
Specific heat capacity (C) of water = 1 Cal /gºC
Change in temperature (ΔT) =.?
Q = MCΔT
20000 = 80 × 1 × ΔT
20000 = 80 × ΔT
Divide both side by 80
ΔT = 20000 / 80
ΔT = 250 °C
Therefore, the change in temperature observed is 250 °C