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Stells [14]
4 years ago
13

How many grams of NaCl are in 150 mL of 3.36 M NaCl

Chemistry
1 answer:
Elenna [48]4 years ago
5 0
I have a solution to a similar problem here but different given values.

Answer:

0.409 g

Explanation:

As stated, 1 mole of NaCl weighs 58.44 g.

A 0.02 M solution contains 0.02 moles per liter.

Hence, 1 litre would contain:

58.44 x 0.02 = 1.1688 g

However, there are only 350 ml, which is 0.35 liters.

Therefore, there would be 0.35 x 1.1688 g in total, 0.409 g

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Based on the periodic table why are be, bg, ca and sr in the same column/group/family?
Lesechka [4]

Answer:

The elements are in the same column/group IIA.

See the explanation below, please.

Explanation:

The elements Calcium, Strontium, Beryllium, Magnesium, Barium and Radio, belong to the group of alkaline earth metals located in group IIA of the periodic table, they require 2 electrons to complete their octet (they have 2 valence electrons). reagents than alkali metals.

7 0
3 years ago
How many grams of CO are produced when 41.0 g of C reacts?
CaHeK987 [17]

Answer:

95.7 g CO to the nearest tenth.

Explanation:

2C + O2 ---> 2CO

Using relative atomic masses:

24 g C produces  2*12 + 2*16 g CO.

So 41 g produces  ( (2*12 + 2*16) * 41  ) / 24

= 95.7 g CO,

7 0
3 years ago
Elements found in the same periods of the modern periodic table are _____.
Dennis_Churaev [7]
Try C I’m sorry if you get it wrong but I’m like 90% sure
5 0
3 years ago
Read 2 more answers
An organic compound composed of carbon and hydrogen connected only by single bonds is an ________. alkane alkene alkyne aromatic
My name is Ann [436]
Alkane, the others have double covalent bonds
4 0
3 years ago
In going from room temperature (25 C) to 10 C above room temperature, the rate of reaction doubles. Calculate the activation ene
Sauron [17]

Answer:

Ea=5.29 × 10⁴ J/mol

Explanation:

In going from 25 °C (298 K) to 35 °C (308 K), the rate of the reaction doubles. Since the rate of the reaction depends on the rate constant (k), this implies that the rate constant doubles. We can find the activation energy (Ea) using the two-point form of the Arrhenius equation.

ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} .(\frac{1}{T_{2}}-\frac{1}{T_{1}})\\ln\frac{2k_{1}}{k_{1}}=\frac{-Ea}{8.314J/K.mol}.(\frac{1}{308K}-\frac{1}{298K} )\\Ea=5.29 \times 10^{4} J/mol

8 0
3 years ago
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