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3 years ago

7. Which part moves the blood away from the heart?

Chemistry

2 answers:

ki77a [65]3 years ago

7 0

Answer:

Veins

Explanation:

Veins, which usually look blue, return blood to the heart.

igor_vitrenko [27]3 years ago

5 0

D. blood vessels, have a good day and make sure u eat!!

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How many valence electrons does Oxygen have:<br> a) 3 <br> b) 5 <br> c) 13

ivolga24 [154]

Answer:

it has 6?

Explanation:

4 0

2 years ago

Read 2 more answers

And with solution...

Tems11 [23]

Answer:

The answer to your question is: V = 6.93 L

Explanation:

Data

N₂ = 5.6 g

Volume of NH₃ = ?

14 g of N ---------------- 1 mol

5.6 g ----------------------- x

x = (5.6 x 1) / 14 = 0.4 mol of N

Reaction

N₂ + 3H₂ ⇒ 2NH₃

1 mol of N₂ ---------------- 2 moles of NH₃

0.4 mol of N₂ -------------- x

x = (0.4 x 2) / 1

x = 0.8 mol of NH₃

Formula

PV = nRT

P = 5200 torr = 6.84 atm

V = ?

n = 0.8

R = 0.082 atm L/ mol °K

T = 450°C = 723°K

Substitution

V = (0.8)(0.082)(723) / 6.84

V = 6.93 L

7 0

3 years ago

Consider the diagram of the Earth, Sun, and Moon system. Which phenomenon is MOST DIRECTLY caused by the revolution of the Earth

zlopas [31]

It is commonly known that our Galaxy rotates upon the sun because of gravity. The diagrams' arrows sort of describe it. Id suggest Gravity as the answer.

5 0

3 years ago

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?

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

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

8 0

3 years ago

The reactivity of metals on the periodic table

kari74 [83]

Explanation:

The most reactive metals are found on the left of the periodic table, in the blue column, known as the alkali metals. Their reactivity increases as we go down column (group) one. Reactive metals, when attached to less reactive metals, have the ability to prevent the less reactive metal from rusting.

8 0

3 years ago