Answer:
Para el CO₂ sabemos que:
densidad = 0,001976 g/cm³
Sabemos que:
densidad = masa/volumen
Entonces, si tenemos una masa de 28 g, podemos escribir:
volumen = masa/densidad
volumen = (28g)/(0,001976 g/cm³) = 14,170 cm^3
Para obtener la masa molar (es decir, la masa de un mol de esta sustancia) simplemente sumamos la masa de un mol de cada componente.
Carbono: tiene una masa molar de 12 g/mol
Oxígeno: tiene una masa molar de 16 g/mol (y tenemos dos oxígenos)
entonces la masa molar va a ser:
masa molar = 12g/mol + 2*16g/mol = 44 g/mol
Es decir, un mol de CO₂, pesa 44 gramos.
Answer:
9: a^8/c^2
Step-by-step explanation:
(a^4/c)^2
(a^8/c^2)
Quantity of gasoline needed by a car to run 800 miles = 30 gallons
Quantity of gasoline needed by a car to run 1 mile =
= 30 ÷ 800
= 0.0375 gallons
So , to run 1 mile a car would need = 0.0375 gallons of oil
To run 700 miles the quantity of gasoline needed =
= 700 × 0.0375
= 26.25 gallons of gasoline
Therefore , a car will use 26.25 gallons of gasoline on a trip of 700 miles .
The three whites:
(30 choose 1) * (29 choose 1) * (28 choose 1) = 30 * 29 * 28
The two reds:
(20 choose 1) * (19 choose 1) = 20 * 19
So, The three whites + two reds = 30 * 29 * 28 + 20 * 19 = 24740
<u>Answer:</u>
Cost of package of paper = 4$
Cost of stapler = 7$
<u>Explanation:</u>
Consider the cost of package of paper = x and that of stapler = y.
Now, we are given that cost of 3 paper packages and 4 staplers = 40$
Hence we get, 3x + 4y = 40 as 1st equation.
we are also given, cost of 5 paper packages and 6 staplers = 62$
Hence, the second equation is 5x + 6y = 62
Now, solving the two equations by method of elimination, we first equate coefficients of any one variable say x by multiplying 1st equation by 5 and second by 3 we get ->
15x + 20y = 200
15x + 18y= 186
Subtracting the two we get y = 7 and substituting this value of y in first equation we get x = 4
which gives the required cost of one paper package = x = 4$
and one stapler = y = 7$
