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3 years ago

How many sucrose molecules will be in 250 grams

Chemistry

1 answer:

sasho [114]3 years ago

6 0

<h2>Answer:</h2>

We will need to know Avogadro's number and the molar mass of sucrose for this problem to do dimensional analysis.

Avogadro's number: 6.022 × 10²³ molecules
Molar mass of sucrose: 342.2965 g/mol

250g × $\frac{1 mol}{342.297g}$ × $\frac{6.022 * 10^{23} molecules}{1 mol}$ = 4.398 molecules

There are <em>4.398 sucrose molecules </em>in 250 grams of sucrose.

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When each of the following pairs of aqueous solutions is mixed, does a precipitation reaction occur? If so, write balanced molec

Aleks04 [339]

Yes, a precipitation process does take place here. BaCO₃ and KOH are the byproducts of this double displacement reaction, and since BaCO₃ is soluble in solution, it precipitates as a solid.

Molecular: Ba(NO₃)₂(aq) + K₂CO₃(aq) → 2K⁺(aq) + 2NO₃¯(aq) + BaCO₃(s) ⬇︎

The complete ionic reaction can be written as follows:

Ba²⁺(aq) + 2NO₃¯(aq) + 2K⁺(aq) + CO₃²¯(aq) → 2K⁺(aq) + 2NO₃¯(aq) + BaCO₃(s) ⬇︎

All the ions in solution have been written out. Now, on to the net ionic equation:

Ba²⁺(aq) + CO₃²¯(aq) → BaCO₃(s) ⬇︎

<h3>what is precipitation?</h3>

Any liquid or frozen water that condenses in the atmosphere and falls to the ground is known as precipitation. Snow, sleet, and rain are just a few of the variations. One of the three main processes that make up the global water cycle is precipitation, along with evaporation and condensation.

Water vapor in the clouds condenses into increasing-sized droplets of water, forming precipitation. Droplets fall to the Earth when they are sufficiently heavy. The water droplets may freeze to produce ice if a cloud is colder, as it would be at higher elevations. Depending on the temperature within the cloud and at the Earth's surface, these ice crystals eventually discharge to the Earth as snow, hail, or rain.

To learn more about precipitation visit:

brainly.com/question/18109776

#SPJ4

6 0

1 year ago

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen

olga55 [171]

Answer: Rate law= $k[A]^1[B]^2$ , order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is $3L^2mol^{-2}s^{-1}$

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[A]^x[B]^y$

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $1.2\times 10^{-2}=k[0.10]^x[0.20]^y$ (1)

From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (2)

Dividing 2 by 1 : $\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}$

$4=2^y,2^2=2^y$ therefore y=2.

b) From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (3)

From trial 3: $9.6\times 10^{-2}=k[0.20]^x[0.40]^y$ (4)

Dividing 4 by 3: $\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}$

$2=2^x,2=2^1$ , x=1

Thus rate law is $Rate=k[A]^1[B]^2$

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1: $1.2\times 10^{-2}=k[0.10]^1[0.20]^2$

$k=3 L^2mol^{-2}s^{-1}$ .

6 0

3 years ago

Define/Differentiate between a physical property and a chemical property?

irga5000 [103]

Answer:

A physical property is an aspect of matter that can be observed or measured without changing it. A chemical property may only be observed by changing the chemical identity of a substance.

Explanation:

i found a table that might be able to help u

4 0

3 years ago

10. At 573K, NO2(g) decomposes forming NO and O2. The decomposition reaction is second order in NO2 with a rate constant of 1.1

leva [86]

Answer:

48.67 seconds

Explanation:

From;

1/[A] = kt + 1/[A]o

[A] = concentration at time t

t= time taken

k= rate constant

[A]o = initial concentration

Since [A] =[A]o - 0.75[A]o

[A] = 0.056 M - 0.042 M

[A] = 0.014 M

1/0.014 = (1.1t) + 1/0.056

71.4 - 17.86 = 1.1t

53.54 = 1.1t

t= 53.54/1.1

t= 48.67 seconds

Hence,it takes 48.67 seconds to decompose.

6 0

2 years ago

Calculate the density of a block of metal with a volume of 12.5 cm cubed and mass of 126.0 g

SVEN [57.7K]

Density is defined as mass/volume (or m/v).

So,

(126.0 g)/(12.5 cm^3)= 10.08 g/cm^3

If your teacher requires correct significant figures, the answer is 10.1 g/cm^3.
If not, the first answer is fine.

3 0

3 years ago

Read 2 more answers