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USPshnik [31]
3 years ago
6

What is the madd in grams of 0.40 moles of sodium borohydride, NaBH4

Chemistry
1 answer:
Orlov [11]3 years ago
3 0

Answer:

Mass = 15.1 g

Explanation:

Given data:

Number of moles of NaBH₄ = 0.40 mol

Mass in gram = ?

Solution:

Formula:

Mass = number of moles × molar mass

Molar mass of NaBH₄ = 37.83 g/mol

By putting values,

Mass = 0.40 mol × 37.83 g/mol

Mass = 15.1 g

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Answer:

The answer to your question is given below.

Explanation:

Rate is simply defined as quantity per unit time. Mathematically it is represented as:

Rate = Quantity /time

Thus, we can obtain the rate as follow:

1. Quantity = 20 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 20/10

Rate = 2 sandwich per mins

2. Quantity = 30 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 30/10

Rate = 3 sandwich per mins

3. Quantity = 40 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 40/10

Rate = 4 sandwich per mins

4. Quantity = 50 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 50/10

Rate = 5 sandwich per mins

Thus, the complete table is given as follow:

Quantity >> Unit of measure >> Rate

10 >>>>>>> 10 min >>>>>>>>>> 1

20 >>>>>>> 10 min >>>>>>>>>> 2

30 >>>>>>> 10 min >>>>>>>>>> 3

40 >>>>>>> 10 min >>>>>>>>>> 4

50 >>>>>>> 10 min >>>>>>>>>> 5

4 0
3 years ago
Is "Hypooxide" even a thing???
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Answer:

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Explanation:

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It has been proposed that North America is moving west 2cm per year. How many kilometers would it move in 5,000 years?
vazorg [7]

The answer is .1 kilometers because 2 times 5,000 equals 10,000 cm.

hope this helps

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3 years ago
HI decomposes to H2 and I2 by the following equation: 2HI(g) → H2(g) + I2(g);Kc = 1.6 × 10−3 at 25∘C If 1.0 M HI is placed into
dsp73

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.037 M

<u>Explanation:</u>

We are given:

Initial concentration of HI = 1.0 M

The given chemical equation follows:

                       2HI(g)\rightleftharpoons H_2(g)+I_2(g)

<u>Initial:</u>               1.0

<u>At eqllm:</u>        1.0-2x          x           x

The expression of K_c for above equation follows:

K_c=\frac{[H_2][I_2]}{[HI]^2}

We are given:

Kc=1.6\times 10^{-3}

Putting values in above expression, we get:

1.6\times 10^{-3}=\frac{x\times x}{(1.0-2x)^2}\\\\x=-0.043,0.037

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.037 M

Hence, the concentration of hydrogen gas at equilibrium is 0.037 M

6 0
3 years ago
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