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3 years ago

13

What is the energy change per gram of ice when an iceberg composed of pure water, cp = 2.06 j/(gk, is heated from -25°c to -15°c

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1 answer:

bonufazy [111]3 years ago

7 0

8 companies i looked it up on google and thats what it said hope its helpful

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What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances

belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope mass amu Relative abundance

1 77.9 14.4

2 81.9 14.3

3 85.9 71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9

% abundance of isotope 1 = 14.4% = $\frac{14.4}{100}=0.144$

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = $\frac{14.3}{100}=0.143$

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = $\frac{71.3}{100}=0.713$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]$

$A=84.2amu$

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

4 0

3 years ago

A 13 g sample of P4010 contains how many

Alex777 [14]

Answer:

$\large \boxed{5.5 \times 10^{22}\text{ molecules of P$_{2}$O}_{5}}$

Explanation:

You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅, then convert to molecules of P₂O₅.

1. Moles of P₄O₁₀

$\text{Moles of P$_{4}$O}_{10} = \text{13 g P$_{4}$O}_{10} \times \dfrac{\text{1 mol P$_{4}$O}_{10}}{\text{283.89 g P$_{4}$O}_{10}} = \text{0.0458 mol P$_{4}$O}_{10}$

2. Moles of P₂O₅

P₄O₁₀ ⟶ 2P₂O₅

The molar ratio is 2 mol P₂O₅:1 mol P₄O₁₀

$\text{Moles of P$_{2}$O}_{5} = \text{0.0458 mol P$_{4}$O}_{10} \times \dfrac{\text{2 mol P$_{2}$O}_{5}}{\text{1 mol P$_{4}$O}_{10}} = \text{0.0916 mol P$_{2}$O}_{5}$

3. Molecules of P₂O₅

$\text{No. of molecules} = \text{0.0916 mol P$_{2}$O}_{5} \times \dfrac{6.022 \times 10^{23}\text{ molecules P$_{2}$O}_{5}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{5.5 \times 10^{22}}\textbf{ molecules P$_{2}$O}_{5}\\\text{There are $\large \boxed{\mathbf{5.5 \times 10^{22}}\textbf{ molecules of P$_{2}$O}_{5}}$}$

6 0

3 years ago

PLEASE ANSWER ASAP I WILL GIVE BRAINLIEST!!

anzhelika [568]

It’s lower, hope this helps with your issue and helps solve your problem, no problem

5 0

2 years ago

Read 2 more answers

How many carbon atoms are found in a molecule with the chemical formula c6h12o6 is probably a carbohydrate and monosaccharide be

valentina_108 [34]

A weirdly worded question. That chemical formula is of glucose which is a carbohydrate as it consists of carbon, oxygen and hydrogen atoms and is a monosaccharide (simple sugar). It had 6 carbon atoms

3 0

3 years ago

A quantity of gas occupies a volume of 804 mL at a temperature of 27 °C. At what temperature will the

irga5000 [103]

Answer:

T₂ = 150 K

Explanation:

Given data:

Initial volume of gas = 804 mL

Initial temperature = 27°C (27+273=300 K)

Final temperature = ?

Final volume = 402 mL

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁

T₂ = 402 mL × 300 K / 804 mL

T₂ = 120,600 mL.K / 804 mL

T₂ = 150 K

3 0

2 years ago