Sean has 5.50 in dimes and quarters. he has 8 more quarters than dimes. how many quarters does he have

___HC2H3O2 + ___KNO2 → ___KC2H3O2 + ___HNO2

Furkat [3]

Let a, b, c, d be coefficients for the balanced reaction:

a HC₂H₃O₂ + b KNO₂ → c KC₂H₃O₂ + d HNO₂

Count the elements among the reactants and products:

H: 4a = 3c + d

C: 2a = 2c

O: 2a + 2b = 2c + 2d

K: b = c

N: b = d

Suppose we let b = 1. Then both c = 1 and d = 1, and in the O equation we have

2a + 2 = 2 + 2 ⇒ 2a = 2 ⇒ a = 1

(Note that the equations for H and C also hold up.)

So the balanced reaction is simply

1 HC₂H₃O₂ + 1 KNO₂ → 1 KC₂H₃O₂ + 1 HNO₂

5 0

2 years ago

Order the following. decimals. from greatest to least. 0.07, 1.06, 0.109, 0.90, 0.601

dangina [55]

1.06 , 0.90, 0.601, 0.109, 0.07

4 0

3 years ago

Read 2 more answers

19x^3 + (14x + 4x^3) =

Brrunno [24]

19x³ + (14x + 4x³)=
You can assume that there is a 1 in front of the parentheses, so you can distribute the one to each term in the parentheses.
19x³ + 1(14x+4x³)=
19x³ + 14x + 4x³=
Then combine like terms to get 23x³ + 14x.
So 19x³ + 14x + 4x³= 23x³ + 14x

6 0

3 years ago

Read 2 more answers

What is the length of side TS?

-Dominant- [34]

We know that

in the triangle TQS
applying the Pythagorean theorem
QS²=TS²+TQ²---------> TQ²=QS²-TS²--------> TQ²=18²-9x²-----> equation 1

in the triangle TRS
TS²=TR²+RS²--------------> TR²=TS²-RS²-------> TR²=9x²-144----> equation 2

in the triangle QTR
TQ²=TR²+36-----------> equation 3

I substitute 1 and 2 in 3
18²-9x²=9x²-144+36--------> 18x²-432=0------> x²=24-------> x=√24
x=2√6
TS=3*x------> 3*2√6-----> 6√6
TS=6√6 units

the answer is
TS=6√6 units
6 is the square root of 6 units

6 0

3 years ago

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Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

3 years ago