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Whitepunk [10]
3 years ago
11

Convert 1.71 × 1024 atoms of carbon to moles of carbon.

Chemistry
1 answer:
beks73 [17]3 years ago
7 0
You need to use Avogadro's constant to convert from atoms of carbon to moles of carbon.

1.71*10^24 atoms C * (1 mole/6.022*10^23 atoms) = 2.84 moles of carbon
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A mixture of oxygen, hydrogen, and nitrogen exerts a total pressure of 378 kPa. If the partial pressures of oxygen and hydrogen
sertanlavr [38]
What's the relationship between total and partial pressure? The total pressure is the sum of the parcial pressures!


So for us, it would be:

378= 212+101+x

where x is the parcial pressure of nitrogen.

Now we count:
378= 212+101+x
378=313+x
378-313=x
65=x

So the parcial pressure exerted by nitrogen is 65!

8 0
3 years ago
SOMEONE PLEASE HELP
Zinaida [17]

Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)

8. 12600 kg/m³ (3 s.f.)

Explanation:

6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.

1 kg= 1000g

1m³= 1 ×10⁶ cm³

Mass of iron bar

= 64.2g

= 64.2 ÷1000 kg

= 0.0642 kg

Volume of iron bar

= 8.16 cm³

= 8.16 ÷ 10⁶

= 8.16 \times 10^{ - 6} \:  kg

\boxed{density =  \frac{mass}{volume} }

Density of iron bar

=  \frac{0.0642}{8.16 \times 10^{ - 6} }

= 7870 kg/m³ (3 s.f.)

7.

\boxed{mass = density \:  \times volume}

Mass

= 1.16 ×28.8

= 33.408 g

= 33.4 g (3 s.f.)

8. Volume of brick

= 12 cm³

= 12  \times  10^{ - 6}  \: m^{3}  \\  = 1.2 \times 10^{ - 5}  \: m ^{3}

Mass of brick

= 151 g

= 151 ÷ 1000 kg

= 0.151 kg

Density of brick

= mass ÷ volume

=  \frac{0.151} {1.25 \times 10^{ - 5} }  \\  = 12600 \: kg/ {m}^{3}

(3 s.f.)

6 0
3 years ago
A 35.0 ml sample of 0.225 m hbr was titrated with 42.3 ml of koh. What is the concentration of the koh?
Lemur [1.5K]

Answer:

The concentration of KOH is 0.186 M

Explanation:

First things first, we need too write out the balanced equation between HBr and KOH.

This is given as;

KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)

From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.

We use the acid base formular in calculating unknown concentrations. This is given as;

\frac{CaVa}{CbVb}  = \frac{na}{nb}

where;

Ca = Concentration of acid

Va = Volume of acid

Cb = Concentration of base

Vb = Volume of base

na = Number of moles of acid

nb = Number of moles of base

KOH is the base and HBr is acid.

Hence;

Ca = 0.225

Va = 35

Cb = ?

Vb = 42.3

na = 1

nb = 1

Making Cb subject of formular we have;

Cb = \frac{CaVaNb}{VbNa}

Cb = (0.225 * 35 * 1) / (42.3 * 1)

Cb = 0.186 M

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3 years ago
Which is an element? ... air ...carbon dioxide ...hydrogen ...water Description
Natalka [10]
All Elements are obtained in the periodic table & water is Not an element it is a compound.
5 0
3 years ago
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