All categories

3 years ago

A solution is prepared by dissolving 51 g of salt in 340 g of

Chemistry

1 answer:

garri49 [273]3 years ago

5 0

Answer:

okķkkkkkkkkkkkkkkkkkkkkk.I don't know the answer. GIVE THANKS

You might be interested in

What is the molar mass of a gas if 0.102 g of the gas occupies 0.070 L at STP? a. 32.6g / mol b. 28.2 g/mol c. 40.0g / mol d. 47

alexgriva [62]

Answer:

A. 32.6 g/mol

Explanation:

First convert the volume of gas to moles using the ratio 1 mol / 22.4 L at STP.

0.070 L • (1 mol / 22.4 L) = 0.00313 mol

Now divide the grams of gas by the moles of gas:

0.102 g / 0.00313 mol = 32.6 g/mol

8 0

3 years ago

If you begin with 1250 grams of N2 and 225 grams of H2 in the reaction that forms ammonia gas (NH3), how much ammonia will be fo

barxatty [35]

N₂+3H₂⇒ 2NH₃
m(NH₃)=1250+225*2=1700 grams
N₂ is the limiting reagent.
1250 grams are left when the maximum amount of ammonia is formed.

8 0

3 years ago

d) A short ton of coal holds about 2.24 x 107 J of energy and costs about $56.45 per short ton. How many Joules of energy in the

Feliz [49]

Answer:

396811.337 J

Explanation:

The cost of one short ton of coal = $56.45

The energy related to the short ton of coal = $2.24\times 10^{7}\ J$

Thus, As according to the question,

$56.45 of coal have $2.24\times 10^{7}\ J$ of energy.

$1 of coal have $\frac{2.24\times 10^{7}}{56.45}\ J$ of energy.

The amount of energy = 396811.337 J

4 0

3 years ago

The conversion of cyclopropane to propene occurs with a first-order rate constant of . How long will it take for the concentrati

Gennadij [26K]

This is an incomplete question, here is a complete question.

The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

Answer : The time taken will be, 17.0 hr

Explanation :

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $2.42\times 10^{-2}\text{ hr}^{-1}$

t = time passed by the sample = ?

a = initial concentration of the reactant = 0.080 M

a - x = concentration left = 0.053 M

Now put all the given values in above equation, we get

$t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}$

$t=17.0\text{ hr}$

Therefore, the time taken will be, 17.0 hr

5 0

3 years ago

PLEASE HURRY!!!

bekas [8.4K]

Answer:

high ph -basic,low pH -acidic

Explanation:

hi I'm from India nice to meet you

5 0

3 years ago