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2 years ago

How many molecules are there in 5H20?

Chemistry

2 answers:

belka [17]2 years ago

8 0

Molar mass of 5H2O = 23.01528 g/mol

Svetach [21]2 years ago

3 0

Answer:

The number before any molecular formula applies to the entire formula. So here you have five molecules of water with two hydrogen atoms and one oxygen atom per molecule. Thus you have ten hydrogen atoms and five oxygen atoms in total.

You might be interested in

What are the correct coefficients when this equation is balanced? K + Br2 KBr

alekssr [168]

2K + Br₂ = 2KBr
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4 0

3 years ago

What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.150 mol of HCl were

irinina [24]

Answer:

pH = 2.21

Explanation:

Hello there!

In this case, according to the reaction between NaF and HCl as the latter is added to the buffer:

$NaF+HCl\rightarrow NaCl+HF$

It is possible for us to see how more HF is formed as HCl is added and therefore, the capacity of this HF/NaF-buffer is diminished as it turns acid. Therefore, it turns out feasible for us to calculate the consumed moles of NaF and the produced moles of HF due to the change in moles induced by HCl:

$n_{HF}^{new}=0.300mol+0.150mol=0.450mol\\\\n_{NaF}^{new}=0.200mol-0.150mol=0.050mol$

Next, we calculate the resulting concentrations to further apply the Henderson-Hasselbach equation:

$[HF]=\frac{0.450mol}{1.0L} =0.450M$

$[NaF]=\frac{0.050mol}{1.0L} =0.050M$

Now, calculated the pKa of HF:

$pKa=-log(6.8x10^{-4})=3.17$

We can proceed to the HH equation:

$pH=pKa+log(\frac{[NaF]}{[HF]} )\\\\pH=3.17+log(\frac{0.05M}{0.45M} )\\\\pH=2.21$

Best regards!

6 0

3 years ago

DUEEE IN 2 minutes!!

Julli [10]

Answer:

A acid. i hope this is right!

Explanation:

4 0

3 years ago

A gas has a volume of 10 L at 200 K and a pressure of 1.5 atm . What would the volume of the gas be at 300 K and 2atm of pressur

Kryger [21]

Answer:

jeyjeyjetnetney

Explanation:

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8 0

3 years ago

The activation energies are 45. 3 kj mol^-1 for k1 and 69. 8 kj mo1-1 for k2. If the rate constants are equal at 320 k, at what

Katyanochek1 [597]

At temperature 298K, the value of k1/k2 will equal to 2.00

According to Arrhenius's equation

Sometimes the Arrhenius equation is written as k = Ae- $^{E/RT}$ , where k is the rate of the chemical reaction, A is a constant that varies depending on the chemicals involved, E is the activation energy, and R is the universal gas constant, and T is the temperature.

$\frac{k1}{k2} = \frac{A_{1} e^{-E_{a1}/ RT} }{A_{2} e^{-E_{a2}/ RT} }$

= $\frac{A1}{A2} e^{(E_{a2}-E_{a1)}/RT }$

Given

Eₐ₂ = 69.8 × 10³ J/mol

Eₐ₁ = 45.3 × 10³ J/mol

R = 8.314J/kmol³

Now if the rate constant is equal means K1/K2 = 1 at T = 320 K

So if we put the values of K1/K2, Ea2,Ea1, T, and R into the formula

On solving

We get

A1/A2 = 1.001 × 10⁻⁴

Now when K1/K2 = 2

Then if we put the value of A1/A2, Ea1, Ea2, R, K1/K2 in the same equation,

On solving we get

T = 298 K

Hence, At temperature 298K k1/k2 will equal to 2.00

Learn more about Arrhenius's equation here brainly.com/question/26724488

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5 0

1 year ago