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AlekseyPX
3 years ago
11

What is the molar mass of Sn(CO3)2?

Chemistry
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

The molar mass and molecular weight of Sn(CO3)2 is 238.728.

Explanation:

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In a neutral atom they are both equal, and their even quantities makes the atom neutral...
6 0
3 years ago
Show work, thanks
GarryVolchara [31]

Answer:

Q.1

Given-

Volume of solution-1 L

Molarity of solution -6M

to find gms of AgNO3-?

Molarity = number of moles of solute/volume of solution in litre

number of moles of solute = 6×1= 6moles

one moles of AgNO3 weighs 169.87 g

so mass of 6 moles of AgNO3 = 169.87×6=1019.22

so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution

7 0
3 years ago
A thermos bottle uses a ______ to keep heat in the thermos
attashe74 [19]

Answer:

a thermos bottle uses a vacuum to keep heat in the thermos.

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2 years ago
The number of kilograms of water in a human body varies directly as the mass of the body. upper a 93​-kg person contains 62 kg o
Bad White [126]

Answer;

56 kg of water


Solution;

Weight = 62 kg of water in 93 kg of a person

W=km

62=93k;

k=2/3

W=(2/3)84

=56 kg

Therefore; a 84-kg person will have 56 kg of water.

4 0
3 years ago
If you have a 1.0 L buffer containing 0.208 M NaHSO3 and 0.134 M Na2SO3, what is the pH of the solution after addition of 50.0 m
Vlada [557]

Answer:

pH = 7.233

Explanation:

Initially, the buffer contains 0.208 moles of NaHSO₃ and 0.134 moles of Na₂SO₃.

NaHSO₃ reacts with NaOH thus:

NaHSO₃ + NaOH → Na₂SO₃ + H₂O

50.0 mL of 1.00 M NaOH are:

0.0500L × (1mol / 1L) = 0.0500moles of NaOH added. That means after the addition are produced  0.0500moles of Na₂SO₃ and consumed 0.0500moles of NaHSO₃. That means final moles of the buffer are:

NaHSO₃: 0.208 mol - 0.050 mol = <em>0.158 mol</em>

Na₂SO₃: 0.134 mol + 0.050 mol = <em>0.184 mol</em>

<em> </em>

As pKa of this buffer is 7.167, it is possible to use H-H equation to find pH, thus:

pH = pKa + log₁₀ [Na₂SO₃] / [NaHSO₃]

pH = 7.167 + log₁₀ [0.184] / [0.158]

<em>pH = 7.233</em>

6 0
3 years ago
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