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PSYCHO15rus [73]
2 years ago
11

A compound contains 29.27% carbon, 51.22% nitrogen, and 19.50% oxygen. What is the molecular formula if the molar mass is 570.5

grams?
Chemistry
1 answer:
sveta [45]2 years ago
3 0

Answer:

its either A or B most likley A if im wrong then im sorry.

Explanation:

have a wonderfull day

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Trying to take the ged test please help me I’m stuck
JulsSmile [24]

Answer: C) Either benzene or oxygen may limit the amount of product that can be formed

Explanation: Benzene and oxygen are the reactants of the equation. What type and the amount of reactants there are in a chemical reaction affects the outcome. Therefore, seeing as benzene and oxygen are the reactants, the answer is C).

3 0
2 years ago
A 58.5 g sample of glass is put into a calorimeter (see sketch at right) that contains 250.0 g of water. The glass sample starts
Ket [755]

Answer:

The specific heat capacity of glass is 0.70J/g°C

Explanation:

Heat lost by glass = heat gained by water

Heat lost by glass = mass × specific heat capacity (c) × (final temperature - initial temperature) = 58.5×c×(91.2 - 21.7) = 4065.75c

Heat gained by water = mass × specific heat capacity × (final temperature - initial temperature) = 250×4.2×(21.7 - 19) = 2835

4065.75c = 2835

c = 2835/4065.75 = 0.70J/g°C

5 0
3 years ago
Read 2 more answers
How many molecules are in 69 grams of Na?
12345 [234]
Molar mass Na = 23.0 g/mol

1 mol ---- 23.0 g
n mol ---- 69 g

n = 69 / 23.0

n = 3.0 moles

1 mole -------- 6.02x10²³ molecules
3.0 moles ---- ?

3.0 *  6.02x10²³ / 1

= 1.806x10²⁴ molecules

hope this helps!




7 0
2 years ago
A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so
Xelga [282]

Answer : The equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

Explanation :

First we have to calculate the initial moles of Fe^{3+} and SCN^-.

\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}

\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol

and,

\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}

\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol

The given balanced chemical reaction is,

Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)

Since 1 mole of Fe^{3+} reacts with 1 mole of SCN^- to give 1 mole of FeSCN^{2+}

The limiting reagent is, SCN^-

So, the number of moles of FeSCN^{2+} = 0.0020 mmole

Now we have to calculate the concentration of FeSCN^{2+}.

\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M

Using Beer-Lambert's law :

A=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution

l = path length

\epsilon = molar absorptivity coefficient

\epsilon and l are same for stock solution and dilute solution. So,

\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}

For trial solution:

The equilibrium concentration of SCN^- is,

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{initial} = 0.00050 M

Now calculate the [FeSCN^{2+}].

C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M

Now calculate the concentration of SCN^-.

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)

[SCN^-]_{eqm}=4.58\times 10^{-8}M

Therefore, the equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

5 0
3 years ago
Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?
chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234  would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample  to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now  we start with 150 grams  of Thorium-234

150 \times \frac{1}{2}=24.1

75 \times \frac{1}{2} =48.2

34.5 \times \frac{1}{2} =72.3

17.25 \times \frac{1}{2} =96.4

8.625\times \frac{1}{2} =120.5

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

A=A_{0} \cdot \frac{1}{2^{n}}

Where

A_0  is the initial sample amount  

n = the number of half-lives that pass in a given period of time.

7 0
3 years ago
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