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bezimeni [28]
3 years ago
13

Convert 1.2030 moles of sodium chloride to grams

Chemistry
2 answers:
vazorg [7]3 years ago
5 0
70.306 would be your answer.
Nikolay [14]3 years ago
4 0
You take the moles and multiply by the molar mass
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Calculate the concentration of the rubidium hydroxide solution in g dm³. (relative atomic mass: Rb = 85; relative formula mass:
Fofino [41]

Answer:

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Explanation:

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Given that 0 find the value of a and the value of b.

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3 years ago
What happens to the buoyant force on an object as it is lowered into water?
PSYCHO15rus [73]
The buoyant force is the upward force or thrust...Simply, consider immersing your hand into a bucket of water. What happens to it? It raises (a little). The raising is due to the buoyant force. Basically it is the upward force/thrust that acts on any object immersed into a fluid. Hence why things float when out in water. Hope this helped!!
8 0
3 years ago
How is phosphorylation of glyceraldehyde 3-phosphate in the payment phase of glycolysis different from phosphorylation of glucos
svet-max [94.6K]

In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place.  During this phase, the conversion of glyceraldehyde-3-phosphate to pyruvate and the coupled formation of ATP take place. Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the payoff phase occurs twice per molecule of glucose.

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The divalent cation also affected the response of the enzyme from the endosperm and shoots to adenine nucleotides and inorganic pyrophosphate.

This phase is also called the glucose activation phase. In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. Steps 1, 2, 3, 4, and 5 together are called the preparatory phase.

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7 0
1 year ago
The solubility of CO2 in water is 0.161 g/100 mL at 20oC and a partial pressure of CO2 of 760 mmHg. What partial pressure of CO2
Schach [20]

<u>Answer:</u> The partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

<u>Explanation:</u>

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

The equation given by Henry's law is:

C_{CO_2}=K_H\times p_{CO_2}       ......(1)

where,

C_{CO_2 = solubility of carbon dioxide in water = 0.161 g/100 mL

K_H = Henry's constant = ?

p_{CO_2} = partial pressure of carbon dioxide = 760 mmHg

Putting values in equation 1, we get:

760mmHg=K_H\times 0.161g/100mL\\\\K_H=\frac{760mmHg}{0.161g/100mL}=4720.5g.mmHg/100mL

Now, calculating the pressure of carbon dioxide using equation 1, we get:

C_{CO_2 = solubility of carbon dioxide in water = 0.886 g/100 mL

K_H = Henry's constant = 4720.5 g.mmHg/100 mL

p_{CO_2} = partial pressure of carbon dioxide = ?

Putting values in equation 1, we get:

p_{CO_2}=4720.5g.mmHg/100mL\times 0.886g/100mL=4182.4mmHg

Hence, the partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

4 0
3 years ago
Consider this equation: 2.524 g (5.1 × 106 g) ÷ (6.85 × 103 g) = ? How many significant figures should the result have?
weqwewe [10]

Answer:

7

Explanation:

2.524g(5.1)(106)g

(6.85)(103)g

=

1364.4744g2

705.55g

=

1364.4744g

705.55

=

1364.4744g

705.55

=1.933916g

6 0
3 years ago
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