How do volcanoes change the Earth's surface?

2 answers:

8 0

Hot ash, gases and molten rock pour out and destroy everything it touches and adding new rock where there was nothing before.

adelina 88 [10]3 years ago

6 0

When a volcano erupts magma comes from the top when the magma cools down it hardens and leaves a hard surface

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For the reaction represented by the equation cl2 + 2kbr ® 2kcl + br2, calculate the percentage yield if 200. g of chlorine react

alexgriva [62]

The % yield if 200g of chlorine react with excess Potassium bromide to produce 410g of bromine is calculated as below

% yield = actual yield/theoretical yield x100
the actual yield = 410 grams

calculate the theoretical yield
by first calculate the moles of chlorine used

mole= mass/molar mass
molar mass of Cl2 = 35.5 x2= 71 g/mol

moles= 200g/71g/mol = 2.82 moles

cl2 +2 KBr = 2KCl +Br2

by use of mole ratio between Cl2 to Br2 which is 1:1 the moles of Br2 is also = 2.82 moles

theoretical mass = moles x molar mass
molar mass of Br2= 79.9 x2= 159.8 g/mol

moles= 2.82g x 159.8 g/mol = 450.64 grams

% yield is therefore = 410g/450.64 x100 = 90.98 %

5 0

3 years ago

Ima 5% glucose solution, how many grams of glucose would be present per 100mL

Sedaia [141]

Percentage Weight-in-volume is defined as the number of grams of a solute in a 100 ml (milliliters) solution.

Percentage Weight-in-volume can tell us about the degree of concentration of a given solution.

The solute can be crystalline or non-crystalline in nature.

The number of grams of glucose present in a 5% glucose solution is 5 grams.

This question is based on a Percentage Weight-in-volume. The formula states that:

a% of a glucose solution = a grams of glucose in a 100 mL solution

Hence, 5% glucose solution = 5 grams of glucose / 100 mL solution

Therefore, the number of grams of glucose present in a 5% glucose solution is 5 grams.

To learn more, visit the link below:

brainly.com/question/8482854

6 0

2 years ago

Two containers, one with a volume of 3.0 L and the other with a volume of 2.0 L contain, respectively, argon gas at 1.1 atm and

aalyn [17]

Answer:

a. $p_T=0.93atm$ .

$p_{Ar}=0.66atm\\\\p_{He}=0.3atm$

$x_{Ar}=0.6875\\\\x_{He}=0.3125$

Explanation:

Hello,

In this case, considering that the valve is opened, we can use the Boyle's law in order to compute the final pressure of argon by considering its initial pressure and volume and a final volume of 5.0 L:

$p_{Ar}=\frac{1.1atm*3.0L}{5.0L}=0.66atm$