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JulijaS [17]
3 years ago
8

Why are some lava's viscous than other's?

Chemistry
1 answer:
Pani-rosa [81]3 years ago
7 0
<span>Viscosity is the resistance to flow (opposite of fluidity). Viscosity depends on primarily on the composition of the magma, and temperature. </span>
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The monoprotic acid from among the following is
oksian1 [2.3K]

Monoprotic acid are acids having only one hydrogen atoms after dissociation into ions from its compound. The monoprotic acid from among the following is HCl. The answer is letter D. HCl → H+ + Cl-. Note that there is only one H+ ion upon dissociation.

3 0
3 years ago
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For the chemical reaction , I identify the reactant and the products .
svetlana [45]

The reactant is Mercury (II) Oxide while the products are Mercury and Oxygen separately.

This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.

Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.

Hope this helps!

7 0
3 years ago
How did Neils Bohr change the model of the atom
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3 years ago
A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch
kozerog [31]

Answer:

\large \boxed{\text{4.63 atm}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ =    1 °C

p₂ = ?;             V₂ = 416 mL; n₂ = n₁; T₂ =  82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (   1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The  new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}

6 0
3 years ago
Mg + _LINO3 → _Mg(NO3)2 + _Li
-BARSIC- [3]

Answer: both numbers are 2

Explanation:

7 0
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