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3 years ago

Which is the most important value of a graph

Chemistry

1 answer:

aksik [14]3 years ago

4 0

X and y axis. i think

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What is the mass of an object that has a density of 2.3 g/mL And a volume of 343ML

Bogdan [553]

Answer:

788.9g

Explanation:

density=m/v

2.3g/ml=m/343ml

=343×2.3

=788.9g

7 0

3 years ago

All the following are signs of a chemical reaction, except

hodyreva [135]

C. a change of state. It can be a physical or a chemical state of change.

8 0

2 years ago

Read 2 more answers

The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-

grin007 [14]

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

$2N_2O \rightarrow 2N_2 + O_2$

rate constant $k = 1.94\times 10^{-4} min^{-1}$

Half life $t_{1/2} = \frac{0.693}{k} = 3572 min$

Initial pressure $N_2 O = 4.70 atm$

Pressure after 3572 min = P

According to first order kinematics

$k = \frac{1}{t} ln\frac{4.70}{P}$

$1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}$

solving for P we get

P = 2.35 atm

$2N_2O \rightarrow 2N_2 + O_2$

initial 4.70 0 0

change -2x +2x +x

final 4.70 -2x 2x x

pressure of $O_2$ after first half life = 2.35 = 4.70 - 2x

x = 1.175

pressure of $N_2$ after first half life = 2x = 2(1.175) = 2.35 ATM

Total pressure = 2.35 + 2.35 + 1.175

= 5.875 atm

5 0

2 years ago

Read 2 more answers

How many grams are in 5.2 grams of K

AleksAgata [21]

I think you meant K as in kilograms?
If so, the answer should be 0.0052.

4 0

2 years ago

Read 2 more answers

A sample was then prepared containing 14.00 mL of coffee and 8.00 mL of 4.80 ppm Li , and diluted to a 50.00 mL total volume. Th

mafiozo [28]

Answer:

The right answer is " $[Na^+]=5.57 \ ppm$ ".

Explanation:

The given values are:

$\frac{A_s}{A_{coffee}} =\frac{0.840}{1.000}$

According to the question,

The concentration of standard will be:

= $\frac{7.80}{50.00}\times 4.80$

= $0.156\times 4.80$

= $0.748 \ ppm$

Coffee after dilution,

⇒ $\frac{A_{coffee}}{C_{coffee}} =F\times \frac{A_s}{C_s}$

or,

⇒ $C_{coffee}=\frac{A_{coffee}}{A_s}\times \frac{C_s}{F}$

On substituting the values, we get

⇒ $=\frac{1.000}{0.840}\times \frac{0.748}{1.68}$

⇒ $=1.191\times 1.3104$

⇒ $=1.561 \ ppm$

hence,

In unknown sample, the concentration of coffee will be:

= $1.561\times \frac{50}{14}$

= $\frac{78.05}{14}$

= $5.57 \ ppm$

7 0

3 years ago