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ZanzabumX [31]
3 years ago
11

What is the mass of 0.50 moles of Na? 11 grams 46 grams 3.0 × 1023 grams 6.0 × 1023 grams

Chemistry
2 answers:
Korolek [52]3 years ago
7 0

n= mass ÷ molar mass

mass= n × molar mass

=0.50 × 23

=11.5 grams

Nikitich [7]3 years ago
7 0

Answer : The mass of 0.50 moles of Na is, 11 grams

Explanation : Given,

Moles of Na = 0.50 mole

Molar mass of Na = 22.99 g/mole

Formula used :

n=\frac{w}{M}

where,

n = number of moles of Na

w = mass of Na

M = molar mass of Na

Now put all the given values in this formula, we get the mass of Na.

0.50mole=\frac{w}{22.99g/mole}

w=11.4g\approx 11g

Therefore, the mass of 0.50 moles of Na is, 11 grams

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What is the percent of oxygen in C6H12O6?
denis-greek [22]

Answer and Explanation:

C6 H12 O6=6×16=96g/mol

96/180×100=53%

4 0
3 years ago
Acid rain damaging marble<br><br> a<br> Chemical Change<br> b<br> Physical Change
Dmitriy789 [7]

Answer:

B

Explanation:

Acid rain

damaging a marble statue or anything else

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8 0
2 years ago
How many milliliters of a 0.211 M HI solution are needed to reduce 24.0 mL of a 0.354 M KMnO4 solution according to the followin
Novay_Z [31]

Answer:

The answer to your question is 242 ml

Explanation:

Data

HI 0.211 M   Volume = x

KMnO₄ 0.354 M   Volume = 24 ml

Balanced Chemical reaction

     12HI + 2KMnO₄ + 2H₂SO₄ → 6I₂ + Mn₂SO₄ + K₂SO₄ + 8H₂O

Process

1.- Calculate the moles of KMnO₄  0.354 M in 24 ml

Molarity = moles / volume (L)

moles = Molarity x volume (L)

moles = 0.354 x 0.024

moles = 0.0085

2.- From the balanced chemical reaction we know that HI and KMnO₄ react in the proportion 12 to 2. Then,

              12 moles of HI --------------- 2 moles of KMnO₄

                x                     --------------- 0.0085 moles of KMnO₄

             x = (0.0085 x 12)/2

             x = 0.051 moles of HI

3.- Calculate the milliliters of HI 0.211 M

Molarity = moles/volume

Volume = moles/molarity

Volume = 0.051/0.211

Volume = 0.242 L or Volume = 242 ml

8 0
3 years ago
When you apply 1000 joules of energy to 50 grams of water its temperature changes to 30 degrees . What was the initial temperatu
expeople1 [14]

Answer:

25.2°C

Explanation:

Given parameters:

Energy applied to the water  = 1000J

Mass of water  = 50g

Final temperature  = 30°C

Unknown:

Initial temperature  = ?

Solution:

To solve this problem, we use the expression below:

            H  = m c Ф

H is the energy absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

     1000  = 50 x 4.184 x (30  -  initial temperature )

     1000  = 209.2(30 - initial temperature)

      4.78  = 30 - initial temperature

      4.78  - 30  = - initial temperature

         Initial temperature  = 25.2°C

7 0
2 years ago
Please Need help asap
ruslelena [56]

Answer:

A

Explanation:

The number of protons and neutrons of an element is the same. the electrons are the only thing that can differ.  The atomic number equal the protons and neutrons.

4 0
2 years ago
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