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jenyasd209 [6]
3 years ago
8

What is the mass of oxygen in 10.0 grams of water?

Chemistry
1 answer:
skelet666 [1.2K]3 years ago
7 0

Answer:

The answer is 16.00 amu.

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12. Classify the following as conduction, convection or radiation. a. A steel mug kept near a stove becomes hot after some time.
Tpy6a [65]

A - convection

B - conduction

C - radiation

D - convection

8 0
3 years ago
Aqueous solutions of barium nitrate and potassium phosphate are mixed. What is the precipitate and how many molecules are formed
kykrilka [37]

Aqueous solutions of barium nitrate and potassium phosphate are mixed. What is the precipitate and how many molecules are formed?

Barium nitrate has a chemical symbol of Ba(NO3)2 and potassium phosphate has a chemical symbol K2PO4. The reaction between these two is a double replacement reaction yielding barium phosphate and potassium nitrate.

The chemical equation representing the reaction is,

            Ba(NO3)2 + K2PO4 à KNO3 + BaPO4

8 0
3 years ago
Read 2 more answers
A sample of sodium-24 with an activity of 14 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24
SOVA2 [1]

Answer:

See explanation

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the sodium-24

t = time taken

Ao = initial activity of sodium-24

A= activity of sodium-24 at time t

a)

0.693/15 = 2.303/15 log (14/A)

0.0462 = 0.1535 log (14/A)

0.0462/0.1535 =  log (14/A)

log (14/A) = 0.0462/0.1535

14/A = Antilog(0.3)

14/A= 1.995

A = 14/1.995

A = 7.0 mCi

b)

0.693/15 = 2.303/30 log (14/A)

0.0462 =0.0768 log(14/A)

0.0462/0.0768 =log (14/A)

(14/A) =Antilog (0.6)

A = 14/Antilog (0.6)

A = 3.5 mCi

c)

0.693/15 = 2.303/45 log (14/A)

0.0462= 0.0512 log (14/A)

log (14/A) = 0.0462/0.0512

log (14/A) = 0.9

(14/A) = Antilog (0.9)

A= 14/Antilog (0.9)

A = 14/7.9

A = 1.77  mCi

d)

2.5 days = 2.5 * 24 hours = 60 hours

0.693/15 = 2.303/60 log (14/A)

0.0462 = 0.03838 log (14/A)

log (14/A) = 0.0462/0.03838

(14/A) = Antilog(1.2)

A= 14/Antilog(1.2)

A = 14/15.8

A = 0.886 mCi

Note that activity (A) decreases as time increases.

5 0
3 years ago
(I)how many atoms are present in 7g of lithium?
ICE Princess25 [194]

Answer :

(i) The number of atoms present in 7 g of lithium are, 6.07\times 10^{23}

(ii) The number of atoms present in 7 g of lithium are, 1.204\times 10^{24}

(iii) The number of moles of F_2 is, 1 mole

The number of moles of CO_2 is, 0.5 mole

The number of moles of OH^- is, 1 mole

Explanation :

<u>Part (i) :</u>

First we have to calculate the moles of lithium.

\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}

Molar mass of Li = 6.94 g/mole

\text{Moles of }Li=\frac{7g}{6.94g/mol}=1.008mole

Now we have to calculate the number of atoms present.

As, 1 mole of lithium contains 6.022\times 10^{23} number of atoms

So, 1.008 mole of lithium contains 1.008\times 6.022\times 10^{23}=6.07\times 10^{23} number of atoms

Thus, the number of atoms present in 7 g of lithium are, 6.07\times 10^{23}

<u>Part (ii) :</u>

First we have to calculate the moles of carbon.

\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}

Molar mass of C = 12 g/mole

\text{Moles of }C=\frac{24g}{12g/mol}=2mole

Now we have to calculate the number of atoms present.

As, 1 mole of carbon contains 6.022\times 10^{23} number of atoms

So, 2 mole of carbon contains 2\times 6.022\times 10^{23}=1.204\times 10^{24} number of atoms

Thus, the number of atoms present in 7 g of lithium are, 1.204\times 10^{24}

<u>Part (iii) :</u>

<u>To calculate the moles of </u>F_2<u> :</u>

\text{Moles of }F_2=\frac{\text{Mass of }F_2}{\text{Molar mass of }F_2}

Molar mass of F_2 = 38 g/mole

\text{Moles of }F_2=\frac{19g}{19g/mol}=1mole

Thus, the number of moles of F_2 is, 1 mole

<u>To calculate the moles of </u>CO_2<u> :</u>

\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}

Molar mass of CO_2 = 44 g/mole

\text{Moles of }CO_2=\frac{22g}{44g/mol}=0.5mole

Thus, the number of moles of CO_2 is, 0.5 mole

<u>To calculate the moles of </u>OH^-<u> ions :</u>

\text{Moles of }OH^-=\frac{\text{Mass of }OH^-}{\text{Molar mass of }OH^-}

Molar mass of OH^- = 17 g/mole

\text{Moles of }OH^-=\frac{17g}{17g/mol}=1mole

Thus, the number of moles of OH^- is, 1 mole

4 0
3 years ago
PH=3 المحلول حمضي او قاعده ​
Dominik [7]

Answer:

去無奈此刻投哦他家

Explanation:

pH=3 المحلول حمضي او قاعده

5 0
2 years ago
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