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mestny [16]
2 years ago
11

Chemical reaction NaCl (aq) + AgNO 3 (aq) → AgCl (s) + NaNO 3 (aq) is:

Chemistry
1 answer:
Vikentia [17]2 years ago
6 0

Answer:

This reaction is both precipitation reaction as well as double displacement reaction.

In a double displacement reaction the anions are exchanged and in a precipitation reaction, a precipitate is formed in the end. In this case, AgCl is the precipitate.

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2) A balloon was inflated to a volume of 5.0 liters at a temperature of
TEA [102]

Answer:

6.12 L

Explanation:

Given that,

Initial volume, V₁ = 5 L

Initial temperature, T₁ = 7.0°C = 343 K

Final temperature, T₂ = 147°C = 420 K

We need to find its new volume. The relation between volume and temperature is given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{5\times 420}{343}\\\\V_2=6.12\ L

So, the new volume is 6.12 L.

8 0
2 years ago
A discus thrower throws a 1.6kg discus at 25m/s what's the kinetic energy?​
Ilya [14]

Answer:

<h2>500 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2} \\

where

m is the mass

v is the velocity

From the question

m = 1.6 kg

v = 25 m/s

We have

k =  \frac{1}{2}  \times 1.6 \times  {25}^{2}  \\  = 0.8 \times 625 \\  = 500

We have the final answer as

<h3>500 J </h3>

Hope this helps you

6 0
3 years ago
Water is Carried to the leaves by stomata
tiny-mole [99]

Answer:

please explain further and i maybe can help you

Explanation:

5 0
2 years ago
PLEASE HELP ME ASAP!
uranmaximum [27]

your answer is b hope this helps


7 0
3 years ago
Read 2 more answers
How many grams are in 1.76 x 10^23 atoms of iodine
Mariana [72]

Answer:

\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

6.022*10^{23}

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Multiply the given number of atoms by Avogadro's number.

1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Flip the fraction so the atoms of iodine will cancel out.

1.76*10^{23} \ atoms \ I*\frac{  1 \ mol \ I}{6.022*10^{23} \ atoms \ I}

1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }

Multiply so the problem condenses into 1 fraction.

\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }

0.2922617071 \ mol \ I

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

  • Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply this fraction by the number of moles found above.

0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply. The moles of iodine will cancel.

0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }

The 1 as a denominator is insignificant.

0.2922617071 *{ 126.9045 \ g\ I }

37.08932581 \ g \ I

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

37.08932581 \ g \ I

The 8 in the hundredth place tells us to round the 0 up to a 1.

\approx 37.1\ g \ I

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

5 0
3 years ago
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