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2 years ago

Someboyd please helppppppp!!!

Chemistry

1 answer:

docker41 [41]2 years ago

3 0

Sodium: Na+

Iron (II): Fe2+

Iron (III): Fe3+

Titanium (IV): Ti4+

Vanadium (V): V5+

Ammonium: NH4+

I hope this helps! :)

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PLEASE HELP What's the mass in grams of 0.442 moles of calcium bromide, CaBr2? The atomic

djverab [1.8K]

Answer: Thus there are 88.4 g of calcium bromide.

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

Molar mass of = 40.1 (1)+ 2(79.9) = 199.9

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

$0.442=\frac{x}{199.9}$

$x=88.4g$

Thus there are 88.4 g of calcium bromide.

7 0

2 years ago

Pa help po science po

mamaluj [8]

Answer:

1. molecule, cell, tissue, organ, organ system, organism, population, community, ecosystem, biosphere

2. A

3. D

4. C

5. C

6. C

7. C

8. B

9. Sorry I don't know :(

10. B

11. A

12. C

13. Sorry :((

14. C

15. a

16. d

Sorry I don't know the rest.. I hope this helps though!!

4 0

1 year ago

With the atomic number 1 and an atomic mass of 1.0079, hydrogen has _________.

GarryVolchara [31]

The answer is A for number 1 and D for 2

4 0

2 years ago

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During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

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3 years ago

What is a picture of homologous pairs of chromosomes and sex chromosomes?

Alecsey [184]

Answer:

allene?

Explanation:

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2 years ago

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