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Setler79 [48]
3 years ago
9

Please help me solve these chemical equations:

Chemistry
2 answers:
Reil [10]3 years ago
8 0
CaCO3 = CaO + CO2
I2 + 2Na2S2O3 = 2NaI + Na2S4O6
6Mg + P4 = 2Mg3P2
2Na + Cl2 = 2NaCl
soldier1979 [14.2K]3 years ago
7 0
This is impossible and confusing!!!
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EASY Chemical reactions<br> Balance and classify
Marat540 [252]

Answer:

double replacement

synthesis

double replacement

i think this one is decomposition

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single replacement

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double replacement

single replacement .....

hopefully i help

6 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
What is 140.2 in scientific notation
Akimi4 [234]

Answer:

1.402 * 10^2.

Explanation:

There are 3 digits before the decimal point so the exponent is 2.

4 0
3 years ago
Read 2 more answers
Assume the recommended single dose of an antihistamine for a child weighing 36–47 lbs is 1.50 teaspoons (tsp) . If 1.00 tsp cont
Scilla [17]

Answer:

18.75 milligrams of active ingredient is the recommended dose for a child weighing 36–47 lbs .

Explanation:

Recommended dose of an antihistamine for a child weighing 36–47 lbs = 1.50 tsp

1 table spoon of antihistamine has 12.5 mg per 5 mL of dose.

Amount of dose in 1 teaspoon = \frac{12.5 mg}{5 mL}=2.5 mg/mL

Volume of 1 tsp = 5 ml

Volume of 1.50 teaspoon = 5 mL\times 1.50 =7.50 mL

Amount of antihistamine in 7.50 mL :

=2.5 mg/mL\times 7.50 mL=18.75 mg

18.75 milligrams of active ingredient is the recommended dose for a child weighing 36–47 lbs .

3 0
3 years ago
In this experiment, 0.070 g of caffeine is dissolved in 4.0mL of water. The caffeine is extracted from the aqueous solution thre
Len [333]

Answer:

0.068 g

Explanation:

The equation that is used to determine the fraction q remaining in water of volume V₁ after n extractions of volume V₂ is:

qⁿ = (V₁/(V₁ + KV₂))ⁿ

Substituting in the values gives the fraction of caffeine left in the water phase:

q³ = (4.0mL/(4.0mL + 4.6(2.0mL))³ = 0.0278

The fraction of caffeine extracted into the methylene chloride phase is:

1 - 0.0278 = 0.972

The amount of caffeine extracted into the methylene chloride is:

(0.070g)(0.972) = 0.068 g

6 0
3 years ago
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