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3 years ago

The atomic radius of a silver atom is 1.44 x 10-10. Give the

Chemistry

1 answer:

professor190 [17]3 years ago

5 0

Answer:

0.144 nm

Explanation:

Silver's electronic configuration is (Kr)(4d)10(5s)1, and it has an atomic radius of 0.144 nm.

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In an electron dot diagram of fluorine , how many dots should be drawn around the elements symbol

Irina18 [472]

Answer:

7 dots

Explanation:

8 0

3 years ago

What is a sign that a chemical change has occurred?

blondinia [14]

Answer:

The production of heat, light, or smoke is observed.

3 0

3 years ago

Read 2 more answers

Given the reaction agi(s) ↔ ag+(aq) + i-(aq) solution equilibrium is reached in the system when

Julli [10]

Missing question:
1) the rate of dissolving reaches zero
<span>2) the rate of crystallization reaches zero </span>
3) the rate of dissolving is zero and the rate of crystallization is greater than zero.
<span>4) both the rate of dissolving and the rate of crystallization are equal and greater than zero.
</span>
Answer is: 4) both the rate of dissolving and the rate of crystallization are equal and greater than zero.
Silver chloride (AgCl) dissolves and form silver and chlorine ions, in the same time silver and chlorine ions crystallizate and form solid salt silver chloride.
In equilibrium rates of dissolvinf and crysallization and concentration of ions do not change.

7 0

3 years ago

An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .

grandymaker [24]

Answer: The specific heat of the unknown metal is $0.897J/g^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= $5040$ Joules

m= mass of substance = 86.8 g

c = specific heat capacity = ?

Initial temperature of the water = $T_i$

Final temperature of the water = $T_f$

Change in temperature , $\Delta T=T_f-T_i=(64.7)^0C$

Putting in the values, we get:

$5040=86.8\times c\times 64.7^0C$

$c=0.897J/g^0C$

The specific heat of the unknown metal is $0.897J/g^0C$

4 0

3 years ago

Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

Gnom [1K]

Answer: The empirical formula for the given compound is $C_5H_7N$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

$C_xH_yN_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of $CO_2=21.363mg=21.363\times 10^3g=21363g$

Mass of $H_2O=6.125g=6.125\times 10^3g=6125g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, $\frac{12}{44}\times 21363=5826.27g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, $\frac{2}{18}\times 6125=680.55$ of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = $\frac{485.52}{97.73}=4.96\approx 5$

For Hydrogen = $\frac{680.55}{97.73}=6.96\approx 7$

For Nitrogen = $\frac{97.73}{97.73}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is $C_5H_7N_1=C_5H_7N$

7 0

2 years ago