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Zielflug [23.3K]
3 years ago
12

Which statement describes the bood type of a person with the alleles /A/B​

Chemistry
1 answer:
ryzh [129]3 years ago
8 0

Answer:

It would be AB

Explanation:

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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
A chunk of sulfur has a volume of 8.33 cm3. What is the mass of this sulfur? (Density of sulfur = 2.07 g/cm3.)
nata0808 [166]

Explanation:

equations to note:

density= mass/volume

mass= volume *density

volume= mass/density

you have a volume- 8.33cm3

you have a density- 2.07 g/cm3

Answer:

8.33cm3 * 2.07g/cm3=  17.24g

mass= 17.24g

5 0
2 years ago
A particular reactant decomposes with a half-life of 113 s when its initial concentration is 0.372 M. The same reactant decompos
AysviL [449]

Answer:

Rate constant =  0.0237 M-1 s-1, Order = Second order

Explanation:

In this problem, it can be observed that as the concentration decreases, the half life increases. This means the concentration of the reactant is inversely proportional to the half life.

The order of reaction that exhibit this relationship is the second order of reaction.

In the second order of reaction, the relationship between rate constant and half life is given as;

t1/2 = 1 / k[A]o

Where;

k = rate constant

[A]o = Initial concentration

k = 1 / t1/2 [A]

Uisng the following values;

k = ?

t1/2 = 113

[A]o = 0.372M

k = 1 / (113)(0.372)

k = 1 / 42.036 = 0.0237 M-1 s-1

8 0
3 years ago
PLEASE HELP!!! WILL MARK BRAINLIEST!!!!!
Hoochie [10]

Answer: 1) D. 2) C. 3) A. 4) C. 5) B. 6) B. 7) A. 8) D. 9) A. 10) C. 11) B.

Explanation: I really hope this helps

4 0
2 years ago
Calculate the number of atoms of carbon in 3.8 moles of methane (CH4)
Harlamova29_29 [7]
There are 6.02*10^23 molecules per mole substances. And there is one carbon atom per molecule of CH4. So the atoms number is 3.8*6.02*10^23=2.29*10^24.
7 0
3 years ago
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