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Whitepunk [10]
3 years ago
12

1s22s22p63s23p64s13d10 which element is this

Chemistry
1 answer:
Sholpan [36]3 years ago
5 0

Answer:

Copper

Explanation:

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a sample of cobalt (specific heat of Co=0.418J/G C )at 100.0 C is dropped into a calorimeter containing 500.0 mL of water at 21.
Alex17521 [72]

Answer:

m_{Co}=6998g=7.0kg

Explanation:

Hello there!

In this case, according to this equilibrium temperature problem, we can set up the following equation to relate the mass, specific heat and temperature change:

Q_{Co}=-Q_{w}\\\\m_{Co}C_{Co}(T_f-T_{Co})=-m_{w}C_{w}(T_f-T_{w})

Thus, we solve for the mass of cobalt as shown below:

m_{Co}=\frac{-m_{w}C_{w}(T_f-T_{w})}{C_{Co}(T_f-T_{Co})} \\\\m_{Co}=\frac{-500.00g*4.184J/g\°C(67.1\°C-21.1\°C)}{0.418J/g\°C(67.1\°C-100\°C)} \\\\m_{Co}=6998g=7.0kg

Best regards!

7 0
3 years ago
Calculate the mass of carbon in 97.0 g of sucrose C12H22O11
lilavasa [31]

1.706515172 ×10 to the power of 23

7 0
3 years ago
Please help quick im on a test and i only got an few mins left
Art [367]

Answer:

hey

Explanation:

7 0
3 years ago
Read 2 more answers
What charged particle is produced during ionic bonds
Setler79 [48]

Answer:

An ionic bond is formed by the complete transfer of some electrons from one atom to another. The atom losing one or more electrons becomes a cation or a positively charged ion. The atom gaining one or more electrons becomes an anion which is a negatively charged ion.

5 0
3 years ago
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X , the freezing point of the solution is
lora16 [44]

Answer:

i=1.62 .

Explanation:

Let, i be the Van't Hoff Factor.

Moles of benzamide,=\dfrac{Mass}{molar\ mass}=\dfrac{70.4}{121.14}=0.58\ mol.

Molality of solution, m=\dfrac{moles  }{mass\ of\ solvent}=\dfrac{0.58}{0.85}=0.68\ molal.

Now, we know

Depression in freezing point, \Delta T=i\times K_f\times m  .....1

It is given that,

\Delta T=2.7^o C\\i=1 ( since\ it\ is\ non\ dissociable\ solutes)\\K_f  ( freezing\ constant)\\

Putting all these values we get,

K_f=3.949\ C/m.

Now, moles of ammonium chloride=\dfrac{70.4}{53.49}=1.316\ mol.

molality =\dfrac{1.316}{0.85}=1.54 molal.\\\Delta T=9.9 .

Putting all these values in eqn 1.

We get,

i=1.62 .

Hence, this is the required solution.

6 0
3 years ago
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